A 2.0-kg block is projected down a plane that makes an angle of 20° with the horizontal with an initial kinetic energy of 2.0 J. If the coefficient of kinetic friction between the block and plane is 0.40, how far will the block slide down the plane before coming to rest?
A) 3.0 m
B) 1.8 m
C) 0.30 m
D) 1.0 m
E) 1.3 m

Question

Quizplus · Accepted Answer

The correct answer is A because the block will slide down the plane until its kinetic energy is converted to thermal energy due to friction. The work done by friction is equal to the change in kinetic energy, so we can use the work-energy theorem to find the distance the block will slide:$$W_{nc}=\Delta K$$$$-f_kd=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$$$-\mu_kmgd\cos	heta=0-2	ext{ J}$$$$d=\frac{2	ext{ J}}{\mu_kmg\cos	heta}=\frac{2	ext{ J}}{(0.40)(2	ext{ kg})(9.8	ext{ m/s}^2)\cos20^\circ}=3.0	ext{ m}$$

A 20-Kg Block Is Projected Down a Plane That Makes an an Angle