A box (M = 20 kg)is sitting on a horizontal surface.It is connected to a massless hook by a light string passing over a massless pulley wheel.The coefficients of friction between the box and the surface are 0.80 (static)and 0.30 (kinetic).If 15 kg (in total)are placed on the hook,calculate the acceleration of the box.
A)10 m/s
B)4.4 m/s
C)2.5 m/s
D)12 m/s
E)0

Question

Quizplus · Accepted Answer

The force of friction acting on the box is:$$f = \mu_s mg = 0.80 	imes 20 	ext{ kg} 	imes 9.8 	ext{ m/s}^2 = 156.8 	ext{ N}$$The tension in the string is:$$T = 15 	ext{ kg} 	imes 9.8 	ext{ m/s}^2 = 147 	ext{ N}$$The net force acting on the box is:$$F = T - f = 147 	ext{ N} - 156.8 	ext{ N} = -9.8 	ext{ N}$$The acceleration of the box is:$$a = F/m = -9.8 	ext{ N} / 20 	ext{ kg} = -0.49 	ext{ m/s}^2$$The negative sign indicates that the box is accelerating to the left.

A Box (M = 20 Kg)is Sitting on a Horizontal