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A 0.50 kg object is at rest. A 2.88 N force to the right acts on the object during a time interval of 1.48 s. a) What is the velocity of the object at the end of this time interval? Answer in units of m/s.

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lublana

Answer:

8.5m/s

Explanation:

We are given that

Mass of object=m=0.50 kg

Initial velocity, u=0

Force=F=2.88 N

Time=1.48 s

a.We know that

[tex]F=ma[/tex]

Using the formula

[tex]2.88=0.50a[/tex]

[tex]a=\frac{2.88}{0.50}=5.76m/s^2[/tex]

[tex]a=\frac{v-u}{t}[/tex]

Using the formula

[tex]5.76=\frac{v-0}{1.48}[/tex]

[tex]v=5.76\times 1.48=8.5m/s[/tex]

Hence, the velocity of the object at the end of this time interval=8.5m/s

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