Question

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

Answer 1

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

$F-mg\sin\theta=ma$

$F-80\sin30=\dfrac{80}{9.8}\times1.5$

$F=\dfrac{80}{9.8}\times1.5+80\sin30$

$F=52\ N$

We need to calculate the work done by the man

Using formula of work done

$W=F\dotc d$

$W=52\times 5.0$

$W=260\ J$

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

Answer 2

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

$\Delta KE = F*d$

Whereas the following is formula for  balance equation:

$F-mg\sin\theta=ma$

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...

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

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