# Mechanics- work energy principle and power

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A golf ball of mass 45.9g is hit from a tee with speed 180km/h . The ball rises to a height of 20m, having traveled along a curved path of length 61.875m. At the highest point of its path the ball is travelling at 144km/h.

a) Find the magnitude of the avg resistance acting on the ball

I calculated this and got resistance = 0.2N

The ball travels a further 105.8m along a curved path to land on the green. The green is 4m lower than the tee. The average resistance remains unchanged.

b) find the speed of the ball just before it lands on the green

Iam getting the ans v= 36m/ s but the textbook ans is 35m/s

m= 0.0459kg, u= 40m/s, s= 105.8- 61.875=43.925m, v=?

Increase in KE = 0.02295v^2-36.72J

Increase in GPE= mgh =-1.836J

Work done against resistance =-8.785J

When I do total mechanical energy = total work done, I get v= 36m/s

a) Find the magnitude of the avg resistance acting on the ball

I calculated this and got resistance = 0.2N

The ball travels a further 105.8m along a curved path to land on the green. The green is 4m lower than the tee. The average resistance remains unchanged.

b) find the speed of the ball just before it lands on the green

Iam getting the ans v= 36m/ s but the textbook ans is 35m/s

m= 0.0459kg, u= 40m/s, s= 105.8- 61.875=43.925m, v=?

Increase in KE = 0.02295v^2-36.72J

Increase in GPE= mgh =-1.836J

Work done against resistance =-8.785J

When I do total mechanical energy = total work done, I get v= 36m/s

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#2

(Original post by

A golf ball of mass 45.9g is hit from a tee with speed 180km/h . The ball rises to a height of 20m, having traveled along a curved path of length 61.875m. At the highest point of its path the ball is travelling at 144km/h.

a) Find the magnitude of the avg resistance acting on the ball

I calculated this and got resistance = 0.2N

The ball travels a further 105.8m along a curved path to land on the green. The green is 4m lower than the tee. The average resistance remains unchanged.

b) find the speed of the ball just before it lands on the green

Iam getting the ans v= 36m/ s but the textbook ans is 35m/s

m= 0.0459kg, u= 40m/s, s= 105.8- 61.875=43.925m, v=?

Increase in KE = 0.02295v^2-36.72J

Increase in GPE= mgh =-1.836J

Work done against resistance =-8.785J

When I do total mechanical energy = total work done, I get v= 36m/s

**Shas72**)A golf ball of mass 45.9g is hit from a tee with speed 180km/h . The ball rises to a height of 20m, having traveled along a curved path of length 61.875m. At the highest point of its path the ball is travelling at 144km/h.

a) Find the magnitude of the avg resistance acting on the ball

I calculated this and got resistance = 0.2N

The ball travels a further 105.8m along a curved path to land on the green. The green is 4m lower than the tee. The average resistance remains unchanged.

b) find the speed of the ball just before it lands on the green

Iam getting the ans v= 36m/ s but the textbook ans is 35m/s

m= 0.0459kg, u= 40m/s, s= 105.8- 61.875=43.925m, v=?

Increase in KE = 0.02295v^2-36.72J

Increase in GPE= mgh =-1.836J

Work done against resistance =-8.785J

When I do total mechanical energy = total work done, I get v= 36m/s

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(Original post by

It would help to see the full working, but for a) I get 0.188N which may account for the later "rounding error"?

**mqb2766**)It would help to see the full working, but for a) I get 0.188N which may account for the later "rounding error"?

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**mqb2766**)

It would help to see the full working, but for a) I get 0.188N which may account for the later "rounding error"?

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#5

(Original post by

The rounding to one decimal place is the ans in the textbook. That's the reason I wrote the resistance as 0.2N

**Shas72**)The rounding to one decimal place is the ans in the textbook. That's the reason I wrote the resistance as 0.2N

Last edited by mqb2766; 1 month ago

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#7

Note I don't understand the s=43? The ball travels a further 105m, so if you're taking ti from the peak, s=105?

But I don't see how you've defined terms (full working).

But I don't see how you've defined terms (full working).

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so m= 0.0459 kg, u= 50m/s, h=20m, s=61.875m, v=144m/s

increase in KE= -20.655J

increase in GPE= mgh = 9.18J

workdone against resistance = -F×S= -61.875F

total mechanical energy = total work done

-20.655+9.18= -61.875F

F=0.185N

increase in KE= -20.655J

increase in GPE= mgh = 9.18J

workdone against resistance = -F×S= -61.875F

total mechanical energy = total work done

-20.655+9.18= -61.875F

F=0.185N

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#10

Guessing you;re taking g=10 rather than my 9.8? Pls upload the full working in one post preferably if you want it checked.

Last edited by mqb2766; 1 month ago

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(Original post by

Guessing you;re taking g=10 rather than my 9.8? Pls upload the full working in one post preferably if you want it checked.

**mqb2766**)Guessing you;re taking g=10 rather than my 9.8? Pls upload the full working in one post preferably if you want it checked.

m=0.0459kg, u= 50m/s, v=?, resistance = 0.2N, s= 105.8m, h=20-4=16m

Increase in KE= 1/2mv^2-1/2mu^2=0.02295v^2-57.375 J

Increase in GPE= mgh= 7.344J

Work done against resistance = -F×S=-21.16J

Total mechanical energy = total work done

I get v= 35 m/s

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**mqb2766**)

Guessing you;re taking g=10 rather than my 9.8? Pls upload the full working in one post preferably if you want it checked.

The ball is travelling vertically when it lands on the green where it is immediately brought to rest.

c) show that the energy absorbed by the green is 30.3J

Can you pls help me with the solution and how to calculate

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#13

(Original post by

Iam stuck with the next question (c)

The ball is travelling vertically when it lands on the green where it is immediately brought to rest.

c) show that the energy absorbed by the green is 30.3J

Can you pls help me with the solution and how to calculate

**Shas72**)Iam stuck with the next question (c)

The ball is travelling vertically when it lands on the green where it is immediately brought to rest.

c) show that the energy absorbed by the green is 30.3J

Can you pls help me with the solution and how to calculate

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#15

(Original post by

mechanical energy

**Shas72**)mechanical energy

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(Original post by

So when the ball comes to rest, what energy is lost by the ball?

**mqb2766**)So when the ball comes to rest, what energy is lost by the ball?

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#17

(Original post by

KE + GPE is lost by the ball

**Shas72**)KE + GPE is lost by the ball

For part b), a sketch would come in handy (pls upload) to make sure the distances, height, signs are correct.

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Iam not getting the ans

I took the KE just before it falls on the ground= -28.11J

final GPE before it falls= -1.836 J taking h=4m

work done against resistance = -0.2×4=-0.8J

so total comes to 30.7J lost by the ball gained by the green.

can you pls help if you can? I have no clue how to get the text book ans 30.3J. pls help

I took the KE just before it falls on the ground= -28.11J

final GPE before it falls= -1.836 J taking h=4m

work done against resistance = -0.2×4=-0.8J

so total comes to 30.7J lost by the ball gained by the green.

can you pls help if you can? I have no clue how to get the text book ans 30.3J. pls help

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#19

(Original post by

Iam not getting the ans

I took the KE just before it falls on the ground= -28.11J

final GPE before it falls= -1.836 J taking h=4m

work done against resistance = -0.2×4=-0.8J

so total comes to 30.7J lost by the ball gained by the green.

can you pls help if you can? I have no clue how to get the text book ans 30.3J. pls help

**Shas72**)Iam not getting the ans

I took the KE just before it falls on the ground= -28.11J

final GPE before it falls= -1.836 J taking h=4m

work done against resistance = -0.2×4=-0.8J

so total comes to 30.7J lost by the ball gained by the green.

can you pls help if you can? I have no clue how to get the text book ans 30.3J. pls help

Last edited by mqb2766; 1 month ago

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I have understood the first and the second question. I don't know how to calculate the energy absorbed by the green . i haven't drawn diagram as I have taken upwards as gain in gpe and loss of KE, downwards there is gain in KE and loss of GPE, height upwards is positive and downwards is negative.

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