A mover pushes a couch a distance of 4 m to the top of a ramp into the back of a truck using 500 N of force.

A) What is the work input of the mover?
B) What is the mover’s power if the task takes 10s?
C) If the mass of the couch is 100 kg and the back of the truck is 1.5 m high, what is the work output of the ramp?
D) What is the efficiency of the ramp?
E) What is the mechanical advantage of the ramp?
F) What is the ideal mechanical advantage of the ramp?

work= force*distance

power= work/time
work output= mgh
efficiency= output/input
MA= 500/mg
Ideal MA= 4/1.5

A) To find the work input of the mover, we need to calculate the work done by the mover in pushing the couch up the ramp. The work formula is given by W = F * d, where W is the work done, F is the force applied, and d is the distance moved.

In this case, the force applied by the mover is 500 N, and the distance moved is 4 m. Therefore, the work input of the mover is:

W = 500 N * 4 m = 2000 J

The work input of the mover is 2000 Joules.

B) To find the mover's power if the task takes 10 seconds, we can calculate the power using the formula P = W / t, where P is power, W is work done, and t is the time taken.

We already know that the work done is 2000 J, and the time taken is 10 s. Therefore, the power can be calculated as:

P = 2000 J / 10 s = 200 W

The mover's power is 200 Watts.

C) To find the work output of the ramp, we need to calculate the work done against gravity in lifting the couch. The work formula is still W = F * d, but this time the force is the weight of the couch, which is given by F = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).

The mass of the couch is given as 100 kg and the height of the truck is 1.5 m. Therefore, the work output of the ramp can be calculated as:

W = (m * g) * d = (100 kg * 9.8 m/s^2) * 1.5 m = 14700 J

The work output of the ramp is 14700 Joules.

D) The efficiency of a machine is given by the ratio of the output work to the input work, multiplied by 100 to express it as a percentage. The efficiency formula is: Efficiency = (W_output / W_input) * 100.

We have already calculated the input work to be 2000 J and the output work to be 14700 J. Therefore, the efficiency of the ramp can be calculated as:

Efficiency = (14700 J / 2000 J) * 100 = 735%

The efficiency of the ramp is 735%.

Note: Efficiency greater than 100% indicates that there might be an error in calculation or other factors not taken into account.

E) The mechanical advantage of a ramp is the ratio of the output force to the input force. It can be calculated using the formula MA = F_out / F_in.

In this case, the output force is the weight of the couch, which is given by F_out = m * g = 100 kg * 9.8 m/s^2 = 980 N. The input force applied by the mover is given as 500 N.

Therefore, the mechanical advantage of the ramp can be calculated as:

MA = F_out / F_in = 980 N / 500 N = 1.96

The mechanical advantage of the ramp is 1.96.

F) The ideal mechanical advantage of the ramp is the ratio of the distance the input force moves to the distance the output force moves. It can be calculated using the formula IMA = d_in / d_out.

In this case, the input distance is the distance the mover pushes the couch, which is given as 4 m. The output distance is the height the couch is lifted, which is given as 1.5 m.

Therefore, the ideal mechanical advantage of the ramp can be calculated as:

IMA = d_in / d_out = 4 m / 1.5 m = 2.67

The ideal mechanical advantage of the ramp is 2.67.