A shopper pushes a grocery cart 20.0 m at constant speed on level
ground, against a 35.0 N frictional force. He pushes in a direction 25.0º
below the horizontal. (a) What is the work done on the cart by friction? (b)
What is the work done on the cart by the gravitational force? (c) What is
the work done on the cart by the shopper? (d) Find the force the shopper
exerts, using energy considerations. (e) What is the total work done on
the guy who answered the first isnt correct
Fx = 35 cos 25
friction work = -35 cos 25 (20)
gravity does none, not in the direction of motion
same as friction work but positive sign
Fx = friction work /20
and
F = Fx/cos 25
total work is friction work since no vertical motion
-34.999
To find the answers to these questions, we will use the work-energy principle and consider the different forces involved. Here are the step-by-step explanations on how to solve each part of the problem.
(a) What is the work done on the cart by friction?
The work done by a force can be calculated using the formula:
Work = Force x Distance x cos(theta),
where theta is the angle between the force and the direction of motion. In this case, the frictional force and the displacement are in opposite directions, so the angle theta is 180 degrees. Therefore, the work done by friction is:
Work_friction = F_friction x d x cos(180) = - F_friction x d
Plugging in the given values, we get:
Work_friction = - 35.0 N x 20.0 m = - 700 J
The negative sign indicates that the friction force is doing negative work, which means it acts in the opposite direction of the displacement.
(b) What is the work done on the cart by the gravitational force?
In this case, the gravitational force is exerted vertically downward, but the displacement is horizontal. Therefore, the angle theta between the force and the displacement is 90 degrees, and cos(theta) will be zero. When cos(theta) is zero, the work done by the force is also zero. So, the gravitational force does no work on the cart.
Work_gravity = 0 J
(c) What is the work done on the cart by the shopper?
The net work done on an object is the sum of the work done by all the external forces acting on it. Since both the frictional force and the gravitational force do work in opposite directions, the net work done on the cart by these forces is the sum of their individual work.
Net work = Work_friction + Work_gravity = -700 J + 0 J = -700 J
The negative sign indicates that the total work done on the cart by these forces is negative, meaning they remove energy from the system.
(d) Find the force the shopper exerts, using energy considerations.
To find the force exerted by the shopper, we need to evaluate the net work done on the cart, and then use the work-energy principle:
Net work = Change in kinetic energy
Since the cart moves at a constant speed, its kinetic energy does not change (no change in speed), so the net work done on the cart is zero. Therefore, the force exerted by the shopper must balance out the work done by the frictional force:
Force_shopper x d x cos(theta) = - F_friction x d
Plugging in the given values, we get:
Force_shopper x 20.0 m x cos(25.0°) = 35.0 N
Solving for Force_shopper, we have:
Force_shopper = (35.0 N) / (20.0 m x cos(25.0°))
Calculating this expression, we find the force exerted by the shopper:
Force_shopper ≈ 42.8 N
(e) What is the total work done on the cart?
The total work done on the cart is the sum of the work done by all the forces acting on it. In this case, since the work done by the gravitational force is zero, the total work done is equal to the work done by friction:
Total work = Work_friction = -700 J