Friction on the Bench and on a Ramp
Students pull a 2.0 kg wooden block across a horizontal laboratory bench with a spring scale. When the block moves at a steady speed, the scale reads 4.9 N. They then place a 3.0 kg metal mass on top and repeat; the block again moves at steady speed and the scale reads 12 N. The bench and block are dry and unchanged. Use g = 9.8 m/s² and treat the scale as horizontal. The students next release the bare 2.0 kg block from rest on a 20° ramp made of the same wood, with no one pulling it. The ramp is long enough to measure motion over 1.0 m. Assume the coefficient of friction on the ramp equals the value found on the bench.
1. Calculate the coefficient of kinetic friction from each trial. Do the two values agree closely enough to treat the wood as one consistent surface? Explain.
Read the worked answer
At steady speed the pulling force equals kinetic friction.
- Trial 1: normal force = 2.0 × 9.8 = 19.6 N.
μ = 4.9 / 19.6 = 0.25. - Trial 2: total mass = 5.0 kg, so normal force = 5.0 × 9.8 = 49 N.
μ = 12 / 49 = 0.2449 ≈ 0.24.
The values differ by only about 0.005, roughly 2%. This is small enough to be explained by scale-reading uncertainty and slight surface variation, so treating the coefficient as about 0.25 is reasonable.
2. Predict the acceleration of the bare block after release on the 20° ramp. Show the force components and state whether it slides down.
Read the worked answer
For the 2.0 kg block on the ramp:
- Down-slope weight component: mg sin 20° = 2.0 × 9.8 × 0.342 ≈ 6.7 N.
- Normal force: mg cos 20° = 2.0 × 9.8 × 0.940 ≈ 18.4 N.
- Kinetic friction up the slope: μN ≈ 0.25 × 18.4 ≈ 4.6 N.
Net force down the slope = 6.7 − 4.6 = 2.1 N.
Acceleration = F/m = 2.1 / 2.0 ≈ 1.1 m/s² down the ramp.
This assumes the block starts sliding; the calculation uses the kinetic coefficient found on the bench.
3. Using the acceleration from the previous task, find the block's speed after traveling 1.0 m from rest and the time taken.
Read the worked answer
Using a ≈ 1.05 m/s²:
- v² = 2as = 2 × 1.05 × 1.0 = 2.10, so v ≈ 1.4 m/s.
- t = v/a ≈ 1.45 / 1.05 ≈ 1.4 s, or t = √(2s/a) ≈ 1.38 s.
So the block reaches about 1.4 m/s after roughly 1.4 s. The exact values depend slightly on rounding, but the method is the key: use the net force to get acceleration, then apply uniformly accelerated motion.