A 4 kg cart rolls at 3 m/s. Over the next stretch of track, +32 J of net work is done on it.
How fast is the cart moving at the end of the stretch?
Well, the theorem does the accounting. The cart already has kinetic energy: ½ × 4 × (3 × 3) is 18 J.
The +32 J of net work raises that store: 18 + 32 = 50, so the final kinetic energy is 50 J.
To finish, run the kinetic energy equation backward, exactly as in the solve lessons: double the 50 J and divide by the 4 kg, so v² = 100 / 4 = 25, and v = 5 m/s.
A 4 kg cart moves at 3 m/s. The net work done on it over a stretch is +32 J. Calculate its speed at the end of the stretch.
Write down the values:
m = 4 kg, v_i = 3 m/s, W_net = +32 J, v_f = ?
Write down the equation:
W_net = KE_f - KE_i
Find the starting store first: KE_i = (0.5 × 4) × (3 × 3), which is 18 J.
Substitute in the values, and calculate:
KE_f = 18 + 32 = 50, then v_f² = 100 / 4 = 25
The cart's final speed is 5 m/s.
A 6 kg sled moves at 2 m/s. The net work done on it over a stretch is +27 J. Enter its kinetic energy at the end of the stretch in J.
The starting store is the whole trap. The net work is the change, never the final amount: a cart already moving keeps its 18 J, and the 32 J lands on top.
A 0.8 kg skateboard rolls at 5 m/s with nobody on it. A shove along its motion does +9.6 J of net work on it. Enter its speed at the end in m/s.
Negative net work runs the same routine downhill: it lowers the store, and the new, smaller store hands back the new, smaller speed.
The same 4 kg cart, now at 5 m/s, has -18 J of net work done on it by friction over the next stretch. Calculate its new speed.
Write down the values:
m = 4 kg, v_i = 5 m/s, W_net = -18 J, v_f = ?
Write down the equation:
W_net = KE_f - KE_i
The starting store is now (0.5 × 4) × (5 × 5), which is 50 J.
Substitute in the values, and calculate:
KE_f = 50 - 18 = 32, then v_f² = 64 / 4 = 16
The cart's new speed is 4 m/s.
An 8 kg crate slides at 6 m/s. Friction does -63 J of net work on it along the floor. Enter its speed at the end in m/s.
An 8 kg delivery cart rolls at 3.5 m/s onto a long straight section where the net force on it stays forward. Over the section, the net work done on the cart is +72 J. Enter the cart's speed at the end of the section in m/s.
A 0.4 kg soccer ball rolls at 6 m/s when a kick does +17 J of net work on it. Enter the ball's speed just after the kick in m/s.
An 800 kg car travels at 12.5 m/s on a test track. As the driver brakes, the forces on the car do -33600 J of net work over a short stretch. Enter the car's speed at the end of the stretch in m/s.