Answer:
60 kg
80 kg
Explanation:
Work is equal to the change in energy.
W = ΔE = E − E₀
Let's start with block B. The work done by the tension force is equal to the change in energy. Initially, the block has potential energy. Finally, the block has kinetic energy.
W = ΔE
FΔy = ½ mv² − mgh
T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)
T (-2.0 m) = m (-2 m²/s²)
T = m (1 m/s²)
Now let's look at block A. The work done by tension and against friction is equal to the change in energy. Initially, the block has no energy. Finally, it has both kinetic and potential energy.
W = ΔE
Fd = ½ mv² + mgh − 0
(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)
(T − Nμ) (2.0 m) = 120 J
T − Nμ = 60 N
Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.
N = mg cos θ
N = (4.00 kg) (10 m/s²) (⅘)
N = 32 N
Substitute:
T − 32μ = 60 N
If μ = 0, then T = 60 N and m = 60 kg.
If μ = ⅝, then T = 80 N and m = 80 kg.
