Answer :
Answer
given,
mass of the box = 9.1 kg
kinetic friction coefficient = 0.375
acceleration = 1.69 m/s²
kinetic frictional force on the box = ?
If the elevator is stationary, the kinetic frictional force is
F = µ m g
F = 0.375 × 9.1 × 9.81
F = 33.48 N.
If the elevator is accelerating upward at 1.88 m/s^2, we add the acceleration to the acceleration due to gravity. The force is
F = µ m (g+a)
F = 0.375 x 9.1 x (9.81 + 1.69)
F = 0.375 x 9.1 x 11.5
F= 42.31 N
If the elevator is accelerating downward, we subtract.
F = µ m (g-a)
F = 0.375 x 9.1 x (9.81 - 1.69)
F = 0.375 x 9.1 x 8.12
F= 27.71 N
Final answer:
The kinetic frictional force on a 9.10-kg box sliding in an elevator is 33.71 N when the elevator is stationary, 38.29 N when accelerating upward at 1.69 m/s², and 29.14 N when accelerating downward at 1.69 m/s².
Explanation:
The kinetic frictional force that acts on a box sliding across the floor of an elevator can be determined using the coefficient of kinetic friction and the normal force acting on the box. The coefficient of kinetic friction (μ_k) is given as 0.375. To find the frictional force (f_k), the equation f_k = μ_k * N is used, where N is the normal force.
- (a) When the elevator is stationary, N equals the weight of the box (mg). Therefore, f_k = 0.375 * 9.10 kg * 9.80 m/s² = 33.71 N.
- (b) When the elevator is accelerating upward, the normal force increases due to the additional force from the acceleration of the elevator. N = m(g + a), so f_k = 0.375 * 9.10 kg * (9.80 + 1.69) m/s² = 38.29 N.
- (c) When the elevator is accelerating downward, the normal force decreases. N = m(g - a), which gives f_k = 0.375 * 9.10 kg * (9.80 - 1.69) m/s² = 29.14 N.
