As 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 weights 25 newtons.
A. Calculate the magnitude of the force of friction between the crate and the floor.
B. The driver of a car made an emergency stop on a straight horizontal road.
C. The wheels locked and the car skidded to a stop. The marks made by the rubber tires on the dry asphalt are 16 meters long, and the car’s mass is 1,200 kilograms.
D. Determine the weight of the car.
E. Calculate the magnitude of the frictional force the road applied to the car stopping it.
The Correct Answer and Explanation is :
Part A: Force of Friction Between the Crate and the Floor
In this situation, the crate is moving to the right and a force of 10 newtons is being exerted on it. The crate weighs 25 newtons, which indicates the normal force acting on it is equal to its weight. The frictional force can be calculated using the relationship between the applied force and the frictional force.
The frictional force, ( f_{\text{friction}} ), is opposing the motion of the crate. Since the crate is already in motion, the frictional force is likely kinetic friction, which is proportional to the normal force.
First, we calculate the coefficient of friction (( \mu_k )) by using the equation for frictional force:
[
f_{\text{friction}} = \mu_k \times N
]
Where:
- ( f_{\text{friction}} ) is the frictional force,
- ( \mu_k ) is the coefficient of kinetic friction,
- ( N ) is the normal force (equal to the weight of the crate, 25 newtons).
Since there is no indication of acceleration, we assume the crate is moving at a constant velocity. This means the net force on the crate is zero, so the applied force is exactly balanced by the frictional force. Therefore:
[
f_{\text{friction}} = 10 \, \text{N}
]
Thus, the magnitude of the frictional force is 10 newtons.
Part B: The Emergency Stop of the Car
When the driver of the car made an emergency stop, the wheels locked, causing the car to skid. This indicates that kinetic friction between the tires and the road surface is the primary force that stopped the car.
Part C: Skid Marks and the Mass of the Car
The car’s mass is given as 1,200 kilograms, and the length of the skid marks is 16 meters. This information will help us in calculating the frictional force needed to stop the car.
Part D: Weight of the Car
The weight of the car is given by the formula:
[
\text{Weight} = m \times g
]
Where:
- ( m ) is the mass of the car (1,200 kg),
- ( g ) is the acceleration due to gravity (( 9.8 \, \text{m/s}^2 )).
So:
[
\text{Weight} = 1200 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 11,760 \, \text{N}
]
Thus, the weight of the car is 11,760 newtons.
Part E: Magnitude of the Frictional Force
The car comes to a stop after skidding a distance of 16 meters. The frictional force is responsible for decelerating the car. We can use the work-energy principle to calculate the frictional force. The work done by the frictional force is equal to the change in kinetic energy of the car.
The kinetic energy of the car when it starts skidding is:
[
KE = \frac{1}{2} m v^2
]
The work done by the frictional force is:
[
W = f_{\text{friction}} \times d
]
Where:
- ( f_{\text{friction}} ) is the frictional force,
- ( d ) is the distance over which the car skidded (16 meters).
Since the car comes to a stop, all of the initial kinetic energy is converted into work done by friction:
[
f_{\text{friction}} \times d = \frac{1}{2} m v^2
]
Solving for ( f_{\text{friction}} ), we need the initial velocity ( v ) of the car, which is not provided. However, we can use the car’s deceleration to relate it to the frictional force. From Newton’s second law:
[
f_{\text{friction}} = \mu_k \times N
]
With further information about the deceleration or velocity, the exact frictional force can be calculated.