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a 0.250 kg cart is being pushed up a track that is inclined at 30° abov…

Question

a 0.250 kg cart is being pushed up a track that is inclined at 30° above the horizontal as shown. friction between the cart and the track is negligible. calculate the minimum power required to push the cart up the track at a constant speed of 2.4 m/s. a 0 w b 0.72 w c 3.0 w d 5.2 w

Explanation:

Step1: Find the force parallel to the incline

When the cart is moving at a constant speed, the net force on it is zero. The force parallel to the incline \(F\) is equal to the component of the gravitational force along the incline. Using \(F = mg\sin\theta\), where \(m = 0.250\space kg\), \(g= 9.8\space m/s^{2}\), and \(\theta = 30^{\circ}\).

$$F=(0.250\space kg)\times(9.8\space m/s^{2})\times\sin30^{\circ}$$
$$F = 0.250\times9.8\times0.5=1.225\space N$$

Step2: Calculate the power

The formula for power is \(P = Fv\). Given \(v = 2.4\space m/s\) and \(F = 1.225\space N\)

$$P=(1.225\space N)\times(2.4\space m/s)=2.94\space W\approx3.0\space W$$

Answer:

C. \(3.0\space W\)