Final answer:
The magnitude of the force exerted on the equipment cart by the teaching assistant is approximately 41.8 N. The work done by this force while the cart moves 28.0 m at a constant velocity is 1064 J. The work done by the force of friction is -1064 J, and the work done by gravity is 0 J, given no vertical displacement.
Step-by-step explanation:
The question involves calculating the work done on an equipment cart considering forces and the displacement involved when a teaching assistant (TA) pushes it. To find the answers, we will apply concepts of physics including Newton's laws, the definition of work, and the components of vectors. Since the cart moves at constant velocity, the net force on the cart is zero, meaning the applied force must balance the frictional force.
(a) Magnitude of the force exerted on the cart by the TA
The horizontal component of the applied force (Fcos(θ)) needs to balance the frictional force, which is 38.0 N in this case:
Fcos(25°) = 38.0 N
This can be solved for F (the magnitude of the force) using the equation:
F = 38.0 N / cos(25°)
F ≈ 41.8 N
(b) Work done by the force the TA exerts on the cart
Work is the product of the force in the direction of the movement and the displacement:
W = Fcos(θ) * d
W = 38.0 N * 28.0 m
W = 1064 J
(c) Work done by the force of friction acting on the cart
Work done by friction is given by:
Wf = -frictional force * displacement
Wf = -38.0 N * 28.0 m
Wf = -1064 J
(d) Work done by the force of gravity acting on the cart
The work done by gravity is zero since there is no vertical displacement:
Wg = m * g * h
Wg = 20.0 kg * 9.8 m/s2 * 0 m
Wg = 0 J