Question

An electric stair lift applies a force of 600 newtons directly upwards, moving the lift at a constant velocity of 0.1 meters per second. The power used by the stairlift is the scalar product of the force and velocity. If the stairs are at an angle of 60 degrees BELOW the HORIZONTAL, how much power is the stairlift using?

Answer 1

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Answer 2

Answer:

HERE IS YOUR EXPLANATION:

Explanation:

Lift applies a →F=F→= 600 N upward force

Stairs are inclined at 60∘60∘ below horizontal

Veclocity of lift is constant, →v=v→= 0.1 ms

The power PP used by the stair-lift is given by the scalar 9dot) product: P=→F⋅→v=|→F|×|→v|×cosθP=F→⋅v→=|F→|×|v→|×cos⁡θ

|→F|=600|F→|=600 and |→v|=0.1|v→|=0.1 are magnitudes of force and velocity, repectively.

θθ is the angle between force and velocity.

The velocity of the lift is directed along the stairs, that is inclined at 60∘60∘ below horizontal, while the force is upwards, along the vertical direction.

The direction 60∘60∘ below the horizontal makes an angle of 30∘30∘ with the vertical, therefore, θ=30∘θ=30∘.

Substituting all the information in the power equation, we get:

P=|→F|×|→v|×cosθ=600×0.1×cos(30∘)=51.9615 WP=|F→|×|v→|×cos⁡θ=600×0.1×cos⁡(30∘)=51.9615 W

Thus, power consumed by the stair-lift is 51.9615 W51.9615 W.

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