Question

A man pulls his luggage on a 30 degree incline with a constant force of 25 N. How
much work does he do after walking 20.0 m?
a) 430 J
b) 330J
c) 1500J
d) 250 J

Answer 1

• Force=F=25N
• Displacement=s=20m
• Angle=30°

We know

$\boxed{\sf W=Fscos\theta}$

$\\ \sf\longmapsto W=25(20)cos30°$

$\\ \sf\longmapsto W=500\times \dfrac{\sqrt{3}}{2}$

$\\ \sf\longmapsto W=250\sqrt{3}$

$\\ \sf\longmapsto W=250(1.73)$

$\\ \sf\longmapsto W\approx 430J$

Answer 2

Option a) 430 J work done he do after walking 20.0 m.

What is work?

Work, in physics, is a measure of energy transfer that happens when an object exists moved over a distance by an external force at least part of which exists involved in the direction of the displacement.

Displacement exists as the distance moved in a straight line, in a provided direction, from the starting point. Displacement exists as a vector quantity as it includes size and direction.

An angle stands the figure formed by two rays, named the sides of the angle, sharing a standard endpoint, named the vertex of the angle.

• Force=F=25N
• Displacement=s=20m
• Angle=30°

⇒$W=Fcos\theta$

⇒$W=25(20)cos 30^{0}$

Then,

⇒$W=50*\sqrt{3} /2$

⇒$W=250\sqrt{3}$

Hence,

⇒$W=250(1.73)$

$W$ approximately 430J.

Hence, Option a) 430 J work done he do after walking 20.0 m.

To learn more about Displacement refer to:

https://brainly.com/question/321442

#SPJ2