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Katyanochek1 [597]
3 years ago
8

Suppose an empty grocery cart rolls downhill in a parking lot. The cart has a maximum speed of 1.3 m/s when it hits the side of

the store and comes to rest 0.30 s later. If an unbalanced force of —65 N stops the cart, what is the mass of the grocery cart?
Physics
1 answer:
Ket [755]3 years ago
4 0

The cart comes to rest from 1.3 m/s in a matter of 0.30 s, so it undergoes an acceleration <em>a</em> of

<em>a</em> = (0 - 1.3 m/s) / (0.30 s)

<em>a</em> ≈ -4.33 m/s²

This acceleration is applied by a force of -65 N, i.e. a force of 65 N that opposes the cart's motion downhill. So the cart has a mass <em>m</em> such that

-65 N = <em>m</em> (-4.33 m/s²)

<em>m</em> = 15 kg

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x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\  v_{o}Sin\alpha=21.828.....(2)

Put the value of v₀ from equation (1) to equation (2)

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\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}

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v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s

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