What force is needed to speed up a frictionless $60 kg$ cart from $4.0 m e t e r s$ per second to $6.5 m e t e r s$ per second in $3.0 s e c o n d s$ ?

$50 N$

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$100 N$

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$5 N$

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$20 N$

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Solution

The correct option is A
$50 N$

Step 1: Given data.
$m = 60 k g, v = 6.5 m / s, u = 4 m / s, t = 3 s$
Step 2: Calculations.
We know that acceleration is given by.
$a = \frac{v - u}{t}$
$a = \frac{6.5 - 4}{3}$
$a = \frac{2.5}{3}$
$a = \frac{25}{30}$
$a = \frac{5}{6} m / s^{2}$
According to Newton's 2nd law of motion.
$F = m \times a$
$F = 60 \times \frac{5}{6}$
$F = 10 \times 5$
$F = 50 N$ .
Hence, option (a) is correct.

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What force is needed to speed up a frictionless 60kgcart from 4.0meters per second to6.5metersper second in3.0seconds ?

The correct option is A 50NStep 1: Given data.m=60kg,v=6.5m/s,u=4m/s,t=3sStep 2: Calculations.We know that acceleration is given by.a=v-uta=6.5-43a=2.53a=2530a=56m/s2According to Newton's 2nd law of motion.F=m×aF=60×56F=10×5F=50N.Hence, option (a) is correct.

What force is needed to speed up a frictionless $60 kg$ cart from $4.0 m e t e r s$ per second to $6.5 m e t e r s$ per second in $3.0 s e c o n d s$ ?