Question

Greezy works in a factory because he does not make enough money from The Charming Beggars. He is pushing a very large crate of 50 kg across the floor at a constant velocity with a force of 200 N. He is pushing on the crate upwards at an angle of 20 degree. See diagram. (a) What is the coefficient of kinetic friction between the ground and the crate? (b) Greezy increases his force to 400 N. What is the acceleration of the crate? Two masses, m_1 = 10 kg and m_2 = 5 kg, are connected by a string that passes over a pulley as shown below. Mass #1 is on a frictional surface. The system is released from rest, After 1.2 s, mass #2 drops 1.0 m. Calculate the coefficient of kinetic friction between the surface and mass #1.

Answer 1

(a) The forces acting on the block is shown in the following figure: F sin 20° 200 Fcos 20° mg From the figure, the normal force acting on block is as follows: Fsin 20° +N -mg = 0 N=mg - F sin 20° = (50 kg)(9.8 m/s2 )–(200 N)sin 20° = 421.6 N From the given data, the block is moving with constant velocity. Hence, the net force acting on the block in horizontal direction is zero. From the Newton's 2nd law, the net force acting on block in horizontal direction is as follows: Fcos 20° - fx=0 F cos 20° – HZN =0

Hence, the coefficient of kinetic friction is as follows: F cos 20°- 4 N = 0 Fcos 20° N (200 N) sin 20° 421. 6N = 0.45] (b) When the Greezy increases the force to 400 N, the normal force acting on the block is as follows: F'sin 20° +N'-mg = 0 N'= mg-F'sin 20° = (50 kg)(9.8 m/s?)–(400 N)sin 20° = 353.2 N From the Newton's 2nd law, the net force acting on block in horizontal direction is as follows: F'cos 20°- fx=ma F' cos 20° – HEN'=ma Hence, the acceleration of the block is as follows: F' cos 20° – 4N' = ma _F'cos 20° – HzN' m (400 N)cos 20°-(0.45)(353.2 N) 50 kg = 4.34 m/s?