A 4.00-kg object has a velocity (5.40 i - 1.20 j) m/s. What is its kinetic energ

Question

A 4.00-kg object has a velocity (5.40 i - 1.20 j) m/s. What is its kinetic energy at this moment? Find the net work done on the object if its velocity changes to (8.00 i + 4.00) m/s. A worker pushing a 35.0-kg wooden crate at a constant speed for 11.6 m along a wood floor does 360 J of work by applying a constant horizontal force of magnitude F on the crate. Determine the value of F. If the worker now applies a force greater than F, describe the subsequent motion of the crate. The crate's speed would remain constant over time. The crate s speed would increase with time. The crate would slow and come to rest. Describe what would happen to the crate if the applied force is less than F. The crate's speed would remain constant over time. The crate would slow and come to rest. The crate's speed would increase with time.

Explanation / Answer

(1)
a)
V = 5.4 i^ - 1.20 j^ m/s
|V| = sqrt(5.4^2 + 1.2^2)
|V| =  5.53 m/s

Kinetic Energy = 1/2*mv^2
Kinetic Energy = 1/2*4*5.53^2 J
Kinetic Energy = 61.2 J

(b)
Vf = 8.0 i^ + 4.0 j^ m/s
|Vf| = sqrt(8^2 + 4^2) m/s
|Vf| = 8.94 m/s

K.Efi = 1/2*mVf^2
K.Efi = 1/2*4.0*8.94^2
K.Efi = 159.85 J

Work Done = Change in Kinetic Energy
Work Done = K.Efin - K.Ein
Work Done = 159.85 - 61.2 J
Work Done = 98.65 J

2
(a)
Work done = Force * Distance
370 = F*11.6
F = 31.9 N

(b)
The Crate's Speed would increase with time.

(c)
The Crate would slow and come to rest. (because of Friction Force.)

(b)

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