A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
1 answer:
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration,
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
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where p is the pressure of the air inside the tires and A is the area of contact between the tire and the car. Therefore, a higher pressure means a larger force F, and eventually if the pressure p is higher enough the force F will be large enough to counterbalance the weight of the car.
In fact, the force applied by the tires upwards to counter-balance the weight of the car (pushing downwards) is
where p is the pressure of the air inside the tires and A is the area of contact between the tire and the car. Therefore, a higher pressure means a larger force F, and eventually if the pressure p is higher enough the force F will be large enough to counterbalance the weight of the car.