A 20 kg child sets off from rest at the top of a playground slide 4 m high.
The slide's surface drags with a 49 N friction force along its whole 6 m length.
Real track, real friction. How fast is the child moving at the bottom?
Friction converts mechanical energy as the child slides: 49 × 6 = 294. So 294 J becomes thermal energy on the way down.
Count the start. Take g = 9.8 N/kg. The gravitational potential energy: (20 × 9.8) × 4 = 784 J. Kinetic energy at the top: 0 J.
Substitute what is known: 0 + 784 = KE_f + 0 + 294.
784 - 294 = 490. The child reaches the bottom with 490 J of kinetic energy.
Solve for the speed: 2 × 490 = 980. 980 ÷ 20 = 49. v = √49 = 7. The child arrives at 7 m/s.
The routine: compute E_thermal from the friction force and the distance it acts through, write the account, and solve the one gap.
Any single unknown works the same way: a landing speed, the thermal energy, or the friction force behind it.
A 20 kg child starts from rest at the top of a slide 4 m high and 6 m long, with a 49 N friction force along the surface. Take g = 9.8 N/kg. Find the child's speed at the bottom.
Write down the values in the question:
m = 20 kg, h = 4 m, friction force 49 N along 6 m; v at the bottom = ?
Compute the thermal energy:
49 × 6 = 294. E_thermal = 294 J.
Compute the starting stores:
20 × 9.8 = 196. 196 × 4 = 784. GPE_i = 784 J; KE_i = 0 J.
Write the account, and solve the gap:
0 + 784 = KE_f + 0 + 294
784 - 294 = 490. KE_f = 490 J.
Solve for the speed:
2 × 490 = 980
980 ÷ 20 = 49
v = √49 = 7 m/s
The child reaches the bottom at 7 m/s.
A friction force of 35 N acts on a toboggan along the full 6 m length of an icy chute. Enter the thermal energy that appears, in J.
A roller coaster car starts from rest holding 882 J of gravitational potential energy above a dip. By the bottom of the dip, 382 J has become thermal energy. Enter the car's kinetic energy at the bottom, in J.
A 10 kg crate slides from rest down a ramp 2.5 m high and 4 m long; friction along the ramp is 30 N. Take g = 9.8 N/kg. Find the crate's speed at the base.
Write down the values in the question:
m = 10 kg, h = 2.5 m, friction force 30 N along 4 m; v at the base = ?
Compute the thermal energy:
30 × 4 = 120. E_thermal = 120 J.
Compute the starting stores:
10 × 9.8 = 98. 98 × 2.5 = 245. GPE_i = 245 J; KE_i = 0 J.
Write the account, and solve the gap:
0 + 245 = KE_f + 0 + 120
245 - 120 = 125. KE_f = 125 J.
Solve for the speed:
2 × 125 = 250
250 ÷ 10 = 25
v = √25 = 5 m/s
The crate reaches the base at 5 m/s.
A 12 kg supply bag slides from rest down a straight delivery chute from 5 m up. Friction along the 8 m chute is 36 N. Take g = 9.8 N/kg. Enter the bag's speed at the bottom, in m/s. Answers within 2% are accepted.
A 50 kg skier sets off from rest at the top of a practice slope 3 m high. The slope's surface is 7 m long, with a 110 N friction force along it. Take g = 9.8 N/kg. Enter her speed at the bottom, in m/s. Answers within 2% are accepted.
An 8 kg mail sack starts 5 m above the bottom of a long spiral mail chute, already moving at 2 m/s. It reaches the bottom moving at 6 m/s. Take g = 9.8 N/kg. Enter the thermal energy that appeared during the slide, in J.
A 40 kg barrel slides from rest down a storage ramp 2 m high and 8 m long into a cellar, arriving at 4 m/s. Take g = 9.8 N/kg. Enter the friction force along the ramp, in N.