A 22 kg box, initially at rest, is pulled by a constant 40 Newton force, as show

Question

A 22 kg box, initially at rest, is pulled by a constant 40 Newton force, as shown in the diagram (the diagram just shows a block with a theta degree of 35). The first 10 m are frictionless and the second 10 m has a kinetic friction coefficient of u = .2.

Using energy: how fast is the box traveling at the end of the first 10 m?

Using energy: how fast is it traveling at the end of the second 10 m?

How hard would you have to pull on the rope (at this angle of 35 degrees) in order to lift the box off the table?

Explanation / Answer

In vertical:

N = m g - F sin35 = 192.66 N

For First 10 m :

Applying work energy theorem,

Work done by force = change in KE

(40 x cos35 x 10) = 22 (v^2 - 0 ) / 2

v = 5.46 m/s ...............Ans

now for next 10m,

f =u N = 38.53 n
Work done by force + work done by friction = change in KE

(40 x cos35 x 10) - (38.53 x 10) = 22( v^2 - 5.46^2) /2

v = 4.96 m/s ........Ans

to lift it of the ground, N = 0

N = m g - F sin35 = 0

F = (22 x 9.8) / sin35 = 376 N ........Ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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