(II) A grocery art with mass of 18 kg is pushed at constant speed along an aisle by a force F = 12 N. The applied force acts at a 20-degree angle to the horizontal. Find the work done by each of the forces on the cart if the aisle is 15 m long.
Please help me solve step by step.
I tried to solve it.
No work done by gravity
No work done by normal
169.14 J by applied force
-169.14J by the friction force
are those correct?
correct.
To solve this problem, we need to calculate the work done by the applied force and the work done by the force of friction.
1. First, let's calculate the work done by the applied force (Fapplied).
The work done by a force can be calculated using the formula: work = force x distance x cosine(theta), where theta is the angle between the force and the direction of motion.
Given:
- Force (Fapplied) = 12 N
- Angle (theta) = 20 degrees
- Distance (d) = 15 m
Plugging in these values into the formula, we have:
work_applied = Fapplied x d x cos(theta)
Since the force and distance are given in the same units (N and m), we can directly substitute the values:
work_applied = 12 N x 15 m x cos(20 degrees)
Calculating this expression, we have:
work_applied ≈ 256.64 J (rounded to two decimal places)
Therefore, the work done by the applied force is approximately 256.64 Joules.
2. Next, let's calculate the work done by the force of friction (Ffriction).
Given that the cart is pushed at a constant speed, we can infer that the force of friction is equal in magnitude but opposite in direction to the applied force. Therefore, the force of friction can be calculated as: Ffriction = -Fapplied.
Substituting the given value:
Ffriction = -12 N
Since the cart is moving at a constant speed, the net work done on it is zero. Therefore, the work done by the force of friction must be equal in magnitude and opposite in sign to the work done by the applied force.
Therefore, the work done by the force of friction is approximately -256.64 J (Note: The negative sign indicates that the work done by the force of friction is in the opposite direction of the displacement).
In summary:
- The work done by the applied force is approximately 256.64 J.
- The work done by the force of friction is approximately -256.64 J.