Question

3) A dock worker pushes a 72 kg crate up a 2.0 m high,

7.0 m long ramp. Ignoring friction, how much work is
done on the crate?

Answer 1

The work done on the 72 kg crate by the dock worker as he pushes it up a 2.0 m high ramp is 1,411.2 joules. The calculation is based on the crate's change in gravitational potential energy, which depends on its mass, the acceleration due to gravity, and the height of the ramp.

Step-by-step explanation:

To calculate the amount of work done on the crate we need to consider the gravitational force acting on it and the vertical displacement. Work is defined as the product of the force along the direction of displacement and the displacement itself. Since we are ignoring friction, the only force doing work against gravity is the weight component along the slope. However, the question simplifies to just considering the vertical displacement since the crate is being lifted against gravity.

The work done on the crate is equivalent to the change in gravitational potential energy, which is given by the formula:

Work = m * g * h

where m is the mass of the crate (72 kg), g is the acceleration due to gravity (9.8 m/s2), and h is the height of the ramp (2.0 m).

By plugging in the numbers:

Work = 72 kg * 9.8 m/s2 * 2.0 m = 1,411.2 joules,Therefore, the work done on the crate is 1,411.2 joules.

Answer 2

Work done on the crate is 1411.2 J

Step-by-step explanation:

Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.

Work done = F * d

where F represents the force and d represents the distance moved by the object.

mass = 72 kg , distance moved by the object is given by 2.0 m

Force F = mass * gravity = 72 * 9.8

= 705.6 N =706 N.

Work done = 706 * 2.0 = 1412 J.