A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30 with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the mand is:A) –200 JB) 61 JC) 140 JD) 200 JE) 260 J

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26-11-2019
Physics

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A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30 with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the mand is:A) –200 JB) 61 JC) 140 JD) 200 JE) 260 J

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-11-27T12:51:40+01:00

Answer:

The work done by the man is 260 J.

(E) is correct option.

Explanation:

Given that,

Force = 80 N

Distance = 5.0 m

Angle = 30°

Acceleration = 1.5 m/s²

Let F be the man's force

We need to calculate the force

Using balance equation

[tex]F-mg\sin\theta=ma[/tex]

[tex]F-80\sin30=\dfrac{80}{9.8}\times1.5[/tex]

[tex]F=\dfrac{80}{9.8}\times1.5+80\sin30[/tex]

[tex]F=52\ N[/tex]

We need to calculate the work done by the man

Using formula of work done

[tex]W=F\dotc d[/tex]

[tex]W=52\times 5.0[/tex]

[tex]W=260\ J[/tex]

Hence, The work done by the man is 260 J.

tifanihayyu tifanihayyu · Answer 2 · 2019-12-04T16:42:41+01:00

If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the man is E) 260 J

Explanation:

Work is the quantity of energy transferred by displacement. The following is formula of work done

[tex]\Delta KE = F*d[/tex]

Whereas the following is formula for balance equation:

[tex]F-mg\sin\theta=ma[/tex]

The equation above is come from Newton's Second Law. It applies to a wide range of physical phenomena, but it is not a fundamental principle like the Conservation Laws and also it applies only if the force is the net external force.

Force = 80 N, therefore
Mass = Force/g = 80 N / 10 = 8 kg
Distance = 5.0 m
Angle = 30°
Acceleration = 1.5 m/s²

For the Case of fictionless slope, the speed at the bottom of a frictionless incline does not depend upon the angle of the incline

Then the work done by the man is...

[tex]w = mgd* sin\theta + \Delta KE\\w = 80 N *5.0 m *sin 30 + (F*d)\\w = 80 N *5.0 m *\frac{1}{2} + (F*d)\\w = 40 N *5.0 m+ (F*d)\\w = 40 N *5.0 m+ (m*a*d)\\w = 200 Nm+ (8*1.5*5)\\w = 200 Nm+ 60\\w = 260 J\\[/tex]

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