Physics test questions

[DannyBoy]

a) It's not accelerating in this case, so the propulsion force equals the frictional force in magnitude, i.e. Work Done = Force x Distance = Force x Velocity x Time = 250 x 20 x 10 = 50 000J

b) The total required force to accelerate a 1000kg mass at (20 - 0)/10 = 2m/s^2 is, Force = Mass x Acceleration = 1000 x 2 =2000N

Total force required = Propulsion force + Frictional force
2000 = F - 250 and therefore F = 2250N

The distance moved is the area under the v-t graph, which is a triangle of base 10 and height 20, Distance = 1/2 x 20 x 10 = 100m

Work Done = Force x Distance = 2250 x 100 = 225000J

c) The acceleration is the same as the last case, 2m/s^2, so force required is 2250N, as before. However, the distance covered is different. Once again, the distance is area under the v-t graph, which is a trapezium this time. Distance = (1/2 x 20 x 10) + (20 x 10) = 300m

Work Done = Force x Distance = 2250 x 300 = 675 000J

EDIT: Beaten by Colin!

[t i l e x]

Thanks to both of you. A harder one this time:

We use an horizontal F force to push a 6kg sled at a constant speed on 4m to the top of a small hill angled at 30 degrees. If uc = 0.2, find the work done on the sled (a) by the F force; (b) by the gravity force; (c) by the friction force. (d) What is the total work done on the sled ?

The answers are: 179J, -118J, -61.4J, 0

When I try to do a force system, it becomes very quickly very messy. I must be doing something wrong

[Alvaro]

You can also work with energies. Whatever work the propulsion force does goes into accelerating the object (kinetic energy) and into heat, because of friction. Using Colin's computations, the distance travelled is 100m.

added_kinetic_energy = 1/2 * 1000 * 20^2 = 200000 J
heat_due_to_friction = 250 * 100 = 25000 J
total_work = 225000 J

EDIT: Ooops! I hadn't seen any of the two previous posts.