In a supermarket parking lot, an employee is pushing ten empty shopping carts, lined up in a straight line. The acceleration of the carts is 0.080 m/s2. The ground is level, and each cart has a mass of 25 kg.
(a) What is the net force acting on any one of the carts?
N
(b) Assuming friction is negligible, what is the force exerted by the fifth cart on the sixth cart?
N
(a) The net force acting on any single cart can be obtained from F = m a
(b) The force acting on the sixth cart must be enough to accelerate carts 6 through 10. Use F = m a again, but the m is defferent.
To calculate the net force acting on any one of the carts, we can use Newton's second law of motion. The formula is:
F = m * a
Where F is the net force, m is the mass of the object, and a is the acceleration.
In this case, the mass of each cart is given as 25 kg. And the acceleration of the carts is given as 0.080 m/s^2.
(a) Calculating the net force acting on any one of the carts:
F = 25 kg * 0.080 m/s^2
F = 2 N
Therefore, the net force acting on any one of the carts is 2 N.
(b) To find the force exerted by the fifth cart on the sixth cart, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Since the sixth cart is being pushed by the fifth cart, the force exerted by the fifth cart on the sixth cart is equal in magnitude but opposite in direction to the force exerted by the sixth cart on the fifth cart.
Therefore, the force exerted by the fifth cart on the sixth cart is also 2 N, but in the opposite direction.