‘A force of 10. newtons toward the right is exerted on a wooden crate initially moving to the right on a horizontal wooden floor: The crate weighs 25 newtons: 121. Calculate the magnitude of the force of friction between the crate and the floor: 122. What is the magnitude of the net force acting on the crate?’
The Correct Answer and Explanation is:
To solve this problem, we need to consider two key forces acting on the crate: the applied force and the frictional force.
1. Force of Friction:
The crate weighs 25 newtons, which means the force due to gravity acting on the crate is 25 N (vertically downward). Since the crate is on a horizontal surface, the normal force (the force exerted by the surface to support the weight of the crate) is equal in magnitude and opposite in direction to the weight of the crate. Therefore, the normal force is also 25 N (vertically upward).
The frictional force can be calculated using the equation:ffriction=μ⋅Nf_{\text{friction}} = \mu \cdot Nffriction=μ⋅N
Where:
- μ\muμ is the coefficient of friction between the crate and the floor.
- NNN is the normal force (which is equal to the weight of the crate, 25 N).
Since the coefficient of friction is not given in the problem, we can only assume that the frictional force is the force that opposes the motion of the crate. We can estimate it based on typical values for static friction between wood and wood, but without more information, we cannot determine its exact value.
2. Magnitude of Net Force:
The net force acting on the crate is the difference between the applied force and the frictional force. The applied force is 10 N to the right, and the frictional force opposes this motion, acting to the left. Thus, the net force FnetF_{\text{net}}Fnet is given by:Fnet=Fapplied−ffrictionF_{\text{net}} = F_{\text{applied}} – f_{\text{friction}}Fnet=Fapplied−ffriction
If we assume that the frictional force is, for example, 5 N (a typical value for kinetic friction), then:Fnet=10 N−5 N=5 NF_{\text{net}} = 10\, \text{N} – 5\, \text{N} = 5\, \text{N}Fnet=10N−5N=5N
Therefore, the magnitude of the net force acting on the crate is 5 N to the right.
Conclusion:
- The magnitude of the force of friction depends on the coefficient of friction, which is not provided. But if we assume it to be 5 N, the frictional force would be 5 N.
- The magnitude of the net force acting on the crate is 5 N (assuming a 5 N frictional force).
This problem involves applying Newton’s Second Law and understanding the relationship between applied forces and friction in the context of motion.
