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LenKa [72]
3 years ago
13

A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with

an initial speed of 1.23 m/s . The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with a coefficient of kinetic friction of 0.255 . How far from the bottom of the chute does the toy zebra come to rest

Physics
2 answers:
Veseljchak [2.6K]3 years ago
6 0

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

            Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

            Em_{f} = K = ½ m v²

            Em₀ = Em_{f}

            ½ m v₀² + m g h = ½ m v²

            v² = v₀² + 2 g h

             

Let's calculate

             v = √(1.23² + 2 9.8 1.69)

             v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

            W = ΔK

            -fr x = 0- ½ m v²  

               

Newton's second law

              N- W = 0

     

The equation for the friction is

               fr = μ N

               fr = μ m g

We replace

             μ m g x = ½ m v²

             x = v² / 2μ g

Let's calculate

            x = 5.89² / (2 0.255 9.8)

            x = 6.94 m

Simora [160]3 years ago
4 0

Answer:

6.45m

Explanation:

See the attachments below

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