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A 125.0 [tex]\(N\)[/tex] carton is pulled up a frictionless baggage ramp inclined at [tex]\(29.0^{\circ}\)[/tex] above the horizontal by a rope exerting a 7 [tex]\(N\)[/tex] pull parallel to the ramp's surface. The carton travels 5.50 [tex]\(m\)[/tex] along the surface of the ramp.

Part A:
Calculate the work done on the carton by the rope.


Sagot :

To calculate the work done on the carton by the rope, we need to use the formula for work:

[tex]\[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \][/tex]

where:
- [tex]\( \text{Force} \)[/tex] is the force applied.
- [tex]\( \text{Distance} \)[/tex] is the distance over which the force is applied.
- [tex]\( \theta \)[/tex] is the angle between the direction of the force and the direction of the movement.

In this scenario:
- The force exerted by the rope, [tex]\( F \)[/tex], is [tex]\( 7 \, \text{N} \)[/tex].
- The distance the carton travels along the ramp, [tex]\( d \)[/tex], is [tex]\( 5.50 \, \text{m} \)[/tex].
- The direction of the force is parallel to the ramp’s surface, which means [tex]\( \theta = 0^\circ \)[/tex].

The cosine of [tex]\(0^\circ\)[/tex] is 1, i.e.,

[tex]\[ \cos(0^\circ) = 1 \][/tex]

So, the equation simplifies to:

[tex]\[ \text{Work} = 7 \, \text{N} \times 5.50 \, \text{m} \times 1 \][/tex]

Now, performing the multiplication:

[tex]\[ \text{Work} = 7 \, \times 5.50 \][/tex]

[tex]\[ \text{Work} = 38.5 \, \text{J} \][/tex]

Therefore, the work done on the carton by the rope is [tex]\( 38.5 \, \text{J} \)[/tex].
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