a nurse pushes a patient on a cart 20.7 m down a hall with a constant speed of 0.88 m/s doing 2000 j of work on the cart. if the cart is released it travels 0.48 m before coming to rest. assuming the nurse exerts a horizontal force on the cart, determine the total weight of the cart plus the patient
To determine the total weight of the cart plus the patient, we can first calculate the force exerted by the nurse on the cart while pushing it down the hall.
The work done by the nurse is given by:
Work = Force x Distance
2000 J = Force x 20.7 m
Force = 2000 J / 20.7 m
Force = 96.62 N
Next, we can calculate the acceleration of the cart when it is released and comes to rest:
Using the equation of motion: v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the cart comes to rest)
u = initial velocity (0.88 m/s)
a = acceleration
s = distance travelled after the nurse stops pushing (0.48 m)
0 = (0.88)^2 + 2a(0.48)
a = -(0.88)^2 / (2*0.48)
a = -0.82 m/s^2
Next, we can calculate the total force acting on the cart from the acceleration when it is released:
F = m*a
Where:
F = Force due to acceleration
m = mass of the cart plus the patient
a = acceleration (-0.82 m/s^2)
Force due to acceleration = m*(-0.82)
Force due to acceleration = 96.62 N
Since the net force acting on the cart when the nurse stops pushing is zero, the total force due to acceleration must be equal to the force exerted while pushing:
m*(-0.82) = 96.62
m = 96.62 / (-0.82)
m ≈ 118.02 kg
Therefore, the total weight of the cart plus the patient is approximately 118.02 kg.