In: Physics
A physics TA pushes a 20.0 kg equipment cart at a constant velocity for a distance of 28.0 m down the hall. The force she exerts on the cart is directed at an angle 20.0° below the horizontal, and the force of friction opposing the motion of the cart is 48.0 N. Determine the following. (a) the magnitude (in N) of the force exerted on the cart by the TA (b) the work done (in J) by the force the TA exerts on the cart (c) the work done (in J) by the force of friction acting on the cart (d) the work done (in J) by the force of gravity acting on the cart
mass of cart, m = 20 kg
distance,s = 28 m
= 20o
frictional force, f = 48 N
Let the force applied by the physics TA be 'F'
a) since the cart moves at a constant velocity its acceleration is zero, hence the net force must be zero,
Component of force along the line of motion is F*cos()
hence,
f = F*cos()
here is the angle between the force and displacement
48 = F*cos(20)
F = 51.08 N
b) Work done by the force is
WF = F*s*cos()
WF = 51.08*28*cos(20) = 1343.985 J
c) Work done by the force of friction
Wf = f*s*cos(180)
since friction acts in opposite direction of motion,i.e, angle between displacement and force of friction is 180o
Wf= 48*28*(-1)
Wf = -1344 J
d) Since force of gravity acts vertically downwards, the angle between force of gravity and displacement is 90o
Wg = mg*s*cos(90) = 0