Question

A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.

Answer 1

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

The given parameters:

• Mass of the box, m = 3.1 kg
• Distance moved by the box, d = 2.5 m
• Coefficient of friction, = 0.35
• Inclination of the force, θ = 30⁰

What is work - done?

• Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

$W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)$

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

$W = (0) d \times cos(30)\\\\W = 0 \ J$

The work done by the force of gravity is calculated as follows;

$W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J$

The work done on the box by the normal force is calculated as follows;

$W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J$

Learn more about work done here: brainly.com/question/8119756