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# some more physics questions watch

1. 1) use the masses and the radii of the earth and moon to determine the weight of a man on the surace of then moon if his weight at the earth's surface is 700N.

mass of earth = 5.98x10^24 kg

2) Determine the force equired to a pull a 0.500kg vlock at a constant speed of 2.5m/s, if the coefficent of friction is 0.17.

3) if a 3kg ball has 400J kinetic energy just before before it strikes the ground, then from what height was it released?

4) calculate the unbalanced force acting on a 5kg cannonball as it accelerates from rest to 150m/s in 0.050s.
2. (Original post by intr3pid)
2) Determine the force equired to a pull a 0.500kg vlock at a constant speed of 2.5m/s, if the coefficent of friction is 0.17.
2) because it is at a constant speed there is no resultant force and teh force pulling it is the same as the friction force which is F=uR (u- coefficient) (R=mg)

F= .17 * (9.81*.5)
F= .17 * 4.905
F= 0.83385 N
3. (Original post by intr3pid)
3) if a 3kg ball has 400J kinetic energy just before before it strikes the ground, then from what height was it released?
0.5*m*V² = 400
V² = 400÷(0.5*m) = 400÷1.5
V² = 266 2/3

V is the speed at just before collision with ground

Using V = √(2gh) h is the height from which it was released

V² = 2gh
V²÷2g = h
h= (266 2/3)÷19.6 = 13.61m
4. (Original post by intr3pid)
4) calculate the unbalanced force acting on a 5kg cannonball as it accelerates from rest to 150m/s in 0.050s.
I think you have to use impulse stuff on this

Ft = m(v-u) u=0 (at rest) v=150m/s m=5Kg t=0.05s
F = (5*(150-0))÷0.05)
F = 750÷0.05
F = 15000 N
5. (Original post by intr3pid)
1) use the masses and the radii of the earth and moon to determine the weight of a man on the surace of then moon if his weight at the earth's surface is 700N.

mass of earth = 5.98x10^24 kg

CAN SUM1 PLEEZ DO THIS 1 IM LOST!!!
6. (Original post by mackin boi)
0.5*m*V² = 400
V² = 400÷(0.5*m) = 400÷1.5
V² = 266 2/3

V is the speed at just before collision with ground

Using V = √(2gh) h is the height from which it was released

V² = 2gh
V²÷2g = h
h= (266 2/3)÷19.6 = 13.61m
i would hav done KE gained = PE lost
so> 400J = m g h
h = 400/(9.8*3) = 13.61 dont understand what u did lol
7. (Original post by mackin boi)
I think you have to use impulse stuff on this

Ft = m(v-u) u=0 (at rest) v=150m/s m=5Kg t=0.05s
F = (5*(150-0))÷0.05)
F = 750÷0.05
F = 15000 N
its strange how different ppl do things ...

i would have done f=ma,
a = 150/0.05 = 3000
f = 5 * 3000 = 15000 N

still we both get the same answer lol

i not sure bout q1 either ill giv it a go on paper first so i dont take ages tryin to type sodding equations (not easy for the new forum helpers among us)
8. (Original post by mackin boi)
CAN SUM1 PLEEZ DO THIS 1 IM LOST!!!
ok here we go: q1

W on earth = mg 700 = m * 9.81 m = 700/9.81 = 71.4kg (3sf)

(using mass of moon = 7.36*10^22kg, r of moon = 1,137.4m > from google )

g = GM/r^2 >>> g = (6.67*10^-11 * 7.36*10^22)/ (1737.4)^2

g on moon = 1.63*10^6 (3sf)

W on moon = m*g = 71.4*(1.63*10^6) = 1.16*10^8N = 116MN.

dunno if thats right, any clever clogs out there are welcome to show us all the right way.
9. (Original post by Mr Tom)
its strange how different ppl do things ...

i would have done f=ma,
a = 150/0.05 = 3000
f = 5 * 3000 = 15000 N

still we both get the same answer lol

i not sure bout q1 either ill giv it a go on paper first so i dont take ages tryin to type sodding equations (not easy for the new forum helpers among us)
Tom u r a don...i just like to complicate things
10. (Original post by Mr Tom)
ok here we go: q1

W on earth = mg 700 = m * 9.81 m = 700/9.81 = 71.4kg (3sf)

(using mass of moon = 7.36*10^22kg, r of moon = 1,137.4m > from google )

g = GM/r^2 >>> g = (6.67*10^-11 * 7.36*10^22)/ (1737.4)^2

g on moon = 1.63*10^6 (3sf)

W on moon = m*g = 71.4*(1.63*10^6) = 1.16*10^8N = 116MN.

dunno if thats right, any clever clogs out there are welcome to show us all the right way.

you could also just use F = GMm/r² (ie the force the moon exerts on the man)

F = (6.67*10-11 * 7.36*1022*71.4)/(1737.4*10³)²
F = 116 N.
11. (Original post by intr3pid)
1) use the masses and the radii of the earth and moon to determine the weight of a man on the surace of then moon if his weight at the earth's surface is 700N.

mass of earth = 5.98x10^24 kg

2) Determine the force equired to a pull a 0.500kg vlock at a constant speed of 2.5m/s, if the coefficent of friction is 0.17.

3) if a 3kg ball has 400J kinetic energy just before before it strikes the ground, then from what height was it released?

4) calculate the unbalanced force acting on a 5kg cannonball as it accelerates from rest to 150m/s in 0.050s.
1. weight at Earth's surface = 700N, mass = 70kg, g on the moon is 1/6 of on Earth, so weight = 700/6 = 117N
2. resistance = 0.17*0.5*10 = 0.85N,
3. 400J gpe = mgh, h = 400/mg = 13 m
4. F=ma = 5*(150/0.05) = 15kN

all done approximately and in a hurry!

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