A 2.0-kg block is projected down a plane that makes an angle of 20$\degree$ 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

Explanation:The work done by friction is:$$W_f=f_kd=\mu_kmgd\sin	heta$$The initial kinetic energy is:$$K_i=\frac{1}{2}mv_i^2$$The final kinetic energy is:$$K_f=0$$The work-energy theorem states that:$$W_{net}=K_f-K_i$$Substituting in the expressions for the work done by friction and the initial and final kinetic energies, we get:$$\mu_kmgd\sin	heta=\frac{1}{2}mv_i^2$$Solving for d, we get:$$d=\frac{v_i^2}{2g\mu_k\sin	heta}$$Substituting in the given values, we get:$$d=\frac{(2.0	ext{ m/s})^2}{2(9.8	ext{ m/s}^2)(0.40)\sin20^\circ}=3.0	ext{ m}$$

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