A 27.0kg block is pulled 4.00m along a smooth horizontal surface

Question

A 27.0kg block is pulled 4.00m along a smooth horizontal surface by an applied horizontal, constant force. While being moved the block accelerates from rest to 1.80m/s.

A. What is the block's change in kinetic energy?

ΔK =  ? J

B. How much work was done on the block by gravity?

W_gravity = ? J

C. How much work was done by the applied force?

W_Fapplied = ? J

D. What is the magnitude of the applied force?

F_applied  = ?N

I do not know how to do this, my professor only touched on the surface of work and energy. I asked him about this problem and he explained to me was the equations to use (like K=1/2 mv^2) but the thing is I don't understand what the parts are (like Ki, F, etc). I need someone to walk me through this please...

Answer 1

The key word "rest" tells us that initial speed v_i = 0.

A. ΔK = K_f - K_i = 1/2 m v_f² - 1/2 m v_i² = 1/2 (27.0 kg) (1.80 m/s)² = 43.7 J [1 J = 1 N•m = 1 kg m²/s²]

B. W_gravity = 0 J because Work = Force • displacement • cos θ, where angle θ is the angle between force and displacement vectors, and the gravitational force is perpendicular to the applied horizontal force.

C. Work due to applied force is the change in K, W_applied = ΔK = 43.7 J.

D. With W_applied also = F_applied • d • cos θ, and θ = 0 because the vectors are in the same direction, then 43.7 J = F_applied • (4.00 m) • (1.00) and F_applied = (43.7 J) / (4.00 m) = 10.9 N.

[I get the same answer by using a kinematic equation to find acceleration a = 0.405 m/s², and F = m•a.]