a shopper in a supermarket pushes a cart with a force of 35 n directed at an angle of 25 downward from the horizontal. the force is just sufficient to over come the various frictional force, so the cart moves at constant speed. what is the net work don't on the cart? why? the shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. if the work done by frictional forces doesn't change, would the shopper's applied forve be larger, smaller, or the same? what about the work done on the cart by the shopper?
Work done by shopper - w- 1586J
To calculate the net work done on the cart, we need to determine the horizontal and vertical components of the applied force and the displacement of the cart.
1. Net work done on the cart:
- The horizontal component of the applied force is given by F_horizontal = F_applied * cos(angle), where F_applied is the applied force and angle is the angle with respect to the horizontal.
- The vertical component of the applied force is given by F_vertical = F_applied * sin(angle).
- Since the cart moves at a constant speed, the net work done on it must be zero. This means that the work done by the applied force is equal in magnitude but opposite in direction to the work done by the frictional forces. Thus, the net work done on the cart is zero.
2. Shopper pushing the cart horizontally in the next aisle:
- The work done by frictional forces doesn't change, which means that the magnitude of the frictional force remains the same.
- As the shopper pushes the cart horizontally, the vertical component of the applied force becomes zero since there is no vertical motion.
- Since the frictional force remains the same and the vertical component of the applied force becomes zero, the magnitudes of the applied force (horizontal component) and the frictional force become equal.
- Therefore, the shopper's applied force would be larger in magnitude to balance the same frictional force.
3. Work done on the cart by the shopper:
- The work done on the cart by the shopper is given by the formula W = F_applied * d * cos(theta), where F_applied is the applied force, d is the displacement, and theta is the angle between the force and displacement vectors.
- As the shopper pushes the cart horizontally, the angle between the applied force and displacement vectors becomes zero degrees (cos(0) = 1), resulting in the maximum magnitude for work done.
- Since the magnitude of the applied force is larger when pushing horizontally, the work done on the cart by the shopper would also be larger.
To determine the net work done on the cart, we need to consider the force applied by the shopper at an angle and the displacement of the cart. The work done by a force is given by the formula:
Work = Force * Displacement * cos(theta)
Where:
- Force is the magnitude of the force applied by the shopper.
- Displacement is the distance the cart moves.
- theta is the angle between the force vector and the displacement vector.
In this case, the force applied by the shopper is 35 N, and the angle is 25 degrees downward from the horizontal. Since the cart moves at constant speed, we know there is no acceleration, which means the net force acting on the cart is zero. Therefore, the force applied by the shopper is balanced by the frictional force.
Since the cart moves at a constant speed, the displacement is equal to the distance covered by the cart. The angle of 25 degrees downward is irrelevant for determining the displacement because it is perpendicular to the direction of motion. Therefore, we can assume the displacement is horizontal.
Now, let's calculate the net work done on the cart:
Work = Force * Displacement * cos(theta)
= 35 N * Displacement * cos(0) [cos(0) because the displacement is horizontal]
Since cos(0) is equal to 1, the equation simplifies to:
Work = 35 N * Displacement
The net work done on the cart is dependent on the distance (displacement) the cart moves horizontally, which has not been given in the question. Therefore, we cannot determine the specific value of the net work done without knowing the displacement.
Moving on to the second part of the question, if the work done by the frictional forces stays the same in the next aisle, it means the frictional force remains constant. However, the shopper is now pushing the cart horizontally.
Since there is no vertical displacement and the cart maintains the same speed, the angle between the applied force and displacement is 0 degrees. The formula for work, as mentioned earlier, is:
Work = Force * Displacement * cos(theta)
= Force * Displacement * cos(0)
= Force * Displacement
As we can see, the angle of 0 degrees also results in cos(0) equaling 1. Therefore, the work done on the cart by the shopper is again dependent on the displacement of the cart. If the frictional force remains the same, the applied force by the shopper would also be the same, as they need to balance the opposing force.