mechanics

hi im stuck on this mechanics question.

a mass of 2.5 kg is released from rest at X and slides down a ramp of height 3m and and width of 4m. when the mass reaches Y at the bottom of the ramp it has a velocity of 5ms-1
the distance between X and Y is 5m.

what is the average frictional force between the mass and the ramp?

1) i worked out the angle and i got 36.87 degrees
2) found the horizontal force of the force which is (sin36.87)x (2.5(9.81))

how can i finish of this question the answer is 8.5N?

Original post by man111111

hi im stuck on this mechanics question.

a mass of 2.5 kg is released from rest at X and slides down a ramp of height 3m and and width of 4m. when the mass reaches Y at the bottom of the ramp it has a velocity of 5ms-1
the distance between X and Y is 5m.

what is the average frictional force between the mass and the ramp?

1) i worked out the angle and i got 36.87 degrees
2) found the horizontal force of the force which is (sin36.87)x (2.5(9.81))

how can i finish of this question the answer is 8.5N?

1. You don't need the angle explicitly, you only need the sine and cosine of it.

\sin \alpha = \dfrac{3}{5}

and

\cos \alpha = \dfrac{4}{5}

.

2. The force you worked out isn't horizontal.

First of all, the particle is going down the ramp at constant acceleration because the coefficient of friction is constant throughout and the only other force acting on the particle is its own weight. This gives us that

ma = mg \sin \alpha - F_{\mathrm{riction}}

resolving parallel to the ramp.
What's

a

? You can use a SUVAT eq. that doesnt include time to determine that.

(edited 5 years ago)

Reply 2

stephen1011

9

Original post by man111111

hi im stuck on this mechanics question.

a mass of 2.5 kg is released from rest at X and slides down a ramp of height 3m and and width of 4m. when the mass reaches Y at the bottom of the ramp it has a velocity of 5ms-1
the distance between X and Y is 5m.

what is the average frictional force between the mass and the ramp?

1) i worked out the angle and i got 36.87 degrees
2) found the horizontal force of the force which is (sin36.87)x (2.5(9.81))

how can i finish of this question the answer is 8.5N?

Solution: https://s18.postimg.cc/5fra7krxl/20180505_183946.jpg

Either the answer you gave is wrong or the information on the question is wrong.. or I am wrong but the answer I got was 4.75 N

Reply 3

man111111

OP

12

Original post by RDKGames

1. You don't need the angle explicitly, you only need the sine and cosine of it.

\sin \alpha = \dfrac{3}{5}

and

\cos \alpha = \dfrac{4}{5}

.

2. The force you worked out isn't horizontal.

First of all, the particle is going down the ramp at constant acceleration because the coefficient of friction is constant throughout and the only other force acting on the particle is its own weight. This gives us that

ma = mg \sin \alpha - F_{\mathrm{riction}}

resolving parallel to the ramp.
What's

a

? You can use a SUVAT eq. that doesnt include time to determine that.

1) i worked out the angle and i got 36.87 degrees
2) found the horizontal force of the force which is (sin36.87)x (2.5(9.81))
3) 14.7- r = (2.5)a
4) worked out acceleration using v^2=u^2+2as and i got a=2.5
5) 14.7- r = (2.5)(2.5)
6) r= 8.45

but i do see your point with not needing the angle. thanks for your trick

Reply 4

Gabriel_L

3

You can use M*G*H = 1/2*M*V^2 F*D work out the gravitational energy take away the kenetic energy at the endthen devide by the length of the ramp

Reply 5

simon0

13

Original post by Gabriel_L

You can use M*G*H = 1/2*M*V^2 F*D work out the gravitational energy take away the kenetic energy at the endthen devide by the length of the ramp

Good point although you are missing an addition sign.

Reply 6

lily.grace.kay

13

Original post by Gabriel_L

You can use M*G*H = 1/2*M*V^2 F*D work out the gravitational energy take away the kenetic energy at the endthen devide by the length of the ramp

I know this is an old post but thanks for this, yours was the only explanation that actually made sense to me !

Reply 7

jb137

9

Original post by man111111

1) i worked out the angle and i got 36.87 degrees
2) found the horizontal force of the force which is (sin36.87)x (2.5(9.81))
3) 14.7- r = (2.5)a
4) worked out acceleration using v^2=u^2+2as and i got a=2.5
5) 14.7- r = (2.5)(2.5)
6) r= 8.45

but i do see your point with not needing the angle. thanks for your trick