Question

d. Later, the bobsled is racing down the track at a speed of 20 m/s. With the team in it, the bobsledhas a total mass of 325 kg. How much work must be done by friction to slow down the bobsled to 17m/s? (3 points)

Answer 1

18038.5 J

Step-by-step explanation:

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Answer 2

18037.5 J

Step-by-step explanation:

The work done by friction is equal to the change in the kinetic energy, so it can be calculated as

`\begin{gathered} W=KE_f-KE_i \\ W=(1)/(2)mv_f^2-(1)/(2)mv_i^2 \end{gathered}`

Where m is the mass, vf is the final velocity and vi is the initial velocity.

Replacing m = 325 kg, vf = 17 m/s and vi = 20 m/s, we get

`\begin{gathered} W=(1)/(2)(325\text{ kg\rparen\lparen17 m/s\rparen}^2-(1)/(2)(325\text{ kg\rparen\lparen20 m/s\rparen}^2 \\ \\ W=46962.5\text{ J - 65000 J} \\ W=-18037.5\text{ J} \end{gathered}`

Therefore, the work done by friction is 18037.5 J against the movements.