Question

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?

Answer 1

Answer:

Change in velocity = $0.38 \times 15=5.7 \ m/s.$

Explanation:

According to equation of motion:

$v_f-v_i=a\times t.$

change in velocity = a \times

It is given that force acts, F=25 N.

Also, mass of skier=65 kg.

By newton's force law:

                    $F=m\times a\\Therefore, a=\dfrac{F}{m}.$

                        $a=\dfrac{25}{65}=0.38\ m/s^2.$

Time under which we have to observe change in velocity, t=15 s.

Change in velocity = $0.38 \times 15=5.7 \ m/s.$

Hence, this is the required solution.

Answer 2

The skier's change in velocity is 375 meter per seconds.

Given the following data:

• Force = 25 Newton

• Mass = 65 Kg

• Time = 15 seconds.

To find the skier's change in velocity, we would use the following formula:

$F = \frac{m(v \; - \; u)}{t}$

Substituting the values into the formula, we have;

$25 = \frac{65(v\; - \; u)}{15}$

Cross-multiplying, we have;

$25$ × $15 = v - u$

Change in velocity (v - u) = 375 meter per seconds.

Therefore, the skier's change in velocity is 375 meter per seconds.

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