A crane operator attaches a cable to a 970-kg shipping container that is at rest

Question

A crane operator attaches a cable to a 970-kg shipping container that is at rest. It lifts the container 40 m upward and places it, at rest, on the deck of a ship.

(a) Work-Kinetic Energy Theorem.What is the change in the container's kinetic energy?

K =  J

How much net work was done on the container?

Wnet =  J

(b) Work done by Conservative Forces.How much work was done on the container by gravity?

Wg =  J

How much did the container's gravitational potential energy change?

Ug =  J

(c) Work done by Nonconservative Forces.How much work was done on the container by the tension in the cable?

WT =  J

How much did the mechanical energy of the container change?

Emech =  J

Explanation / Answer

K = 0 becuase it is intially was in rest and finally it was in rest .

Wnet = change in potential energy = m*g*h = 970*9.81*40 = 380628 J

or 380.63 K J

Wg = - 380.63 K J work done by gravity will be same as work done by crane by because gravity act downward and container   move upward so gravity force is opposite to displacement so work doen is negetive .

Ug =   Wg = 380.63 KJ

WT = 380.63 KJ

Emech = ( KE+PE)f  - ( KE+PE)i

0 + m*g*h - (0+0) = mgh = 970 *9.81* 40 = 380.63 KJ

if you have any doubt let me know

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.