A box drops down from a small height onto a horizontal conveyor belt. The mass of the box is 95 kg. The belt moves with speed 1,4 m/s. The coefficient of sliding friction between the box and the belt is 0,42. How long does it take for the box to come to rest relative to the conveyor belt? A4

0,34s

To find the time it takes for the box to come to rest relative to the conveyor belt, we need to consider the forces acting on the box and apply Newton's second law of motion.

First, let's consider the forces acting on the box:
1. The gravitational force (weight) pulling the box downwards, which can be calculated using the formula W = mg, where m is the mass of the box (95 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
W = 95 kg × 9.8 m/s² = 931 N

2. The frictional force opposing the motion of the box on the conveyor belt, which can be calculated using the formula f = μN, where μ is the coefficient of sliding friction (0.42) and N is the normal force. In this case, the normal force is equal to the weight of the box as there is no vertical motion. Therefore, the frictional force is:
f = 0.42 × 931 N ≈ 391 N

Now, let's find the net force acting on the box:
Since the box is coming to rest relative to the conveyor belt, the net force must be equal to zero. Thus, the net force acting on the box is the sum of the gravitational force and the frictional force, but in the opposite direction of motion:
Net force = gravitational force - frictional force
0 = 931 N - 391 N
540 N = 391 N

Now, we can set up the equation using Newton's second law of motion:
Net force = mass × acceleration

Since the net force acting on the box is zero when it comes to rest, the equation becomes:
0 = mass × acceleration
0 = 95 kg × acceleration

Solving for acceleration, we find that the acceleration of the box is zero. This means the box is at rest relative to the conveyor belt.

Therefore, it takes zero seconds for the box to come to rest relative to the conveyor belt since there is no acceleration involved.