Question Video: Power Output Required to Push an Object Upward along a Frictionless Slope at a Particular Speed | Nagwa Question Video: Power Output Required to Push an Object Upward along a Frictionless Slope at a Particular Speed | Nagwa

Question Video: Power Output Required to Push an Object Upward along a Frictionless Slope at a Particular Speed

A cyclist in a race must climb a hill sloping at 7.5° above the horizontal. The cyclist moves with a speed of 5.3 m/s. The mass of the cyclist and their bike is a total of 85 kg. What power output is required from the cyclist?

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Video Transcript

A cyclist in a race must climb a hill sloping at 7.5 degrees above the horizontal. The cyclist moves with a speed of 5.3 meters per second. The mass of the cyclist and their bike is a total of 85 kilograms. What power output is required from the cyclist?

We can call this power output 𝑃 and start off by drawing a diagram of this scenario. In this scenario, a cyclist climbs a hill inclined at an angle we’ve called 𝜃 of 7.5 degrees above the horizontal. The cyclist moves along the ground at a speed 𝑣 of 5.3 seconds and along with their bike has a mass 𝑚 of 85 kilograms. We can recall that power is equal to work done over the time it takes to do that work. In this case, the cyclist is doing work against gravity by moving uphill. Recalling also that work is equal to force done multiplied by distance traveled. If ℎ is the height that the cyclist ascends against gravity, then we can say that the work done is equal to the force of gravity, the cyclist mass times 𝑔, multiplied by ℎ, where 𝑔 is a constant value of 9.8 meters per second squared.

Writing out our equation for power 𝑃, work over time, we can say that our time interval is one second. That means ℎ is the vertical distance the cyclist ascends in one second. Because we know the cyclist’s speed is 5.3 meters per second, that means our right triangle we’ll use to solve for ℎ has a hypotenuse of 5.3 meters. We can write ℎ, therefore, as 5.3 meters times the sin of 𝜃, where 𝜃 is 7.5 degrees. Plugging this expression in for ℎ in our equation for power, we now have an expression for power where all the variables are known or can be calculated.

Plugging in for 𝑚 and 𝑔, when we enter this expression on our calculator, to two significant figures, we find it’s 580 watts. That’s the power output needed from the cyclist to climb this hill, maintaining a speed of 5.3 meters per second.

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