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3 years ago

A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow. What is the skier’s change in velocity

Physics

1 answer:

nikdorinn [45]3 years ago

4 0

Answer:

$\Delta v=5.77m/s$

Explanation:

Newton's 2nd Law relates the net force F on an object of mass m with the acceleration a it experiments by F=ma. In our case the net force is the friction force, since it's the only one the skier is experimenting horizontally and the vertical ones cancel out since he's not moving in that direction. Our acceleration then will be:

$a=\frac{F}{m}$

Also, acceleration is defined by the change of velocity $\Delta v$ in a given time t, so we have:

$a=\frac{\Delta v}{t}$

Since we want the change in velocity, mixing both equations we conclude that:

$\Delta v=at=\frac{Ft}{m}$

Which for our values means:

$\Delta v=\frac{Ft}{m}=\frac{(25N)(15s)}{(65Kg)}=5.77m/s$

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(ii) N = R = Normal Reaction from the ground upwards
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x = X
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$a \: = \: \frac{dv}{dt} \: = \: \frac{25 \: \frac{m}{s} }{50 \: s} \: = \: 0.5 \: \frac{m}{ {s}^{2} }$
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Anna11 [10]

Answer:

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let's start with the swimmer going downstream

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This must be the answer since the return time is known. If you want to delete this time

t_{in}= D / $v_{sg2}$

we substitute

t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

t_{out} = $\frac{D}{v_s +v_r}$

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