A 100 kg mass is pulled along a frictionless surface by a horizontal force F suc

Question

A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 7.0 m/s2. A 20 kg mass slides along the top of the 100 kg mass and has an acceleration of 3.0 m/s2. (It thus slides backward relative to the 100 kg mass.)

(a) What is the frictional force exerted by the 100 kg mass on the 20 kg mass?

60N

(b) What is the net force acting on the 100 kg mass?

N

What is the force F?

N (to the right)

(c) After the 20 kg mass falls off the 100 kg mass, what is the acceleration of the 100 kg mass? (Assume that the force F does not change.)

m/s2 (to the right)

Explanation / Answer

Let me show you the normal way to show exponents here. 7.0 m/s^2 where ^ is shift+6 I assume the frictionless surface is horizontal. a) The 20 kg mass was stationary with respect to the frictionless surface. When the 100 kg mass starts accelerating, the 20 kg mass accelerates in the same direction, but only at 1.0 m/s^2. (You know it's the same direction because both accelerations have are positive.) So there must have been a frictional force, F20, applied to the 20 kg mass. Newton's 2nd Law will give you the magnitude of the force. F20 = m*a I don't know if the force F was to the right or the left. The force on the small mass will be same as on the big one. b) Since the large mass accelerated at 7 m/s^2, the net force that acted on it must have been (using Newton's 2nd again) Fnet = m*a where the values of m and a are now different from what you used above -- different mass and different acceleration. The large mass exerts a force, F20, on the small one. By Newton's 3rd Law, the small mass must have exerted an EQUAL AND OPPOSITE REACTION FORCE on the big one. The net force acting on the 100 kg mass would be the difference between force F and the reaction to the force of friction. Fnet = F + the REACTION to the force of friction (which I called F20) where the reaction force has a negative sign because it points in the OPPOSITE direction. You calculated the value of Fnet above, plug it in and solve for F. c) You now know the value of F. Use it, the mass of the big mass, and Newton's 2nd

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