Question

A net force of 6.6N east acts on a 9.0kg object. If this object accelerates uniformly from rest to a velocity of 3.0m/s east. How far did the object travel and how long did it take?

Answer 1

Solution :-

As per the given data ,

• Force(F) = 6.6 N
• Mass of the object (m)= 9 kg
• Initial velocity (u) = 0 m/s
• Final velocity (v) = 3 m/s

By using Newton's second law of motion ,

⇒ F = ma

On rearranging ,

⇒ a = F / m

⇒ a = 6.6 / 9

⇒ a = 0.73 m/s²

As the object is moving with uniform acceleration we can apply the the first equation of motion in order to find time and second equation of motion in order to find distance

As per the first equation of motion ,

⇒ v = u + at

⇒ 3 = 0 + 0.73 x t

⇒ t = 3 / 0.73

⇒ t = 4.1 s

The time taken by the object is 4.1 s

As per the second equation of motion ,

⇒ s = ut + at² / 2

⇒ s = at² / 2

⇒ s = 0.73 x 4 x 4 / 2

⇒ s = 0.73 x 8

⇒ s = 5.8 m

The distance traveled by the object is 5 .8 m

Answer 2

Explanation:

Given:-

m=9kg

Force=f=6.6N

initial velocity =u=0m/s

final velocity =v=3m/s

To find:-

Distance =s

time taken=t

Solution:-

• According to Eizak Newton's second law of motion.

${:}\longrightarrow$${\boxed{Force {}_{(F)}={Mass} _{(m)}×{acceleration}_{(a)}}}$

${:}\longrightarrow$$6.6=9×a$

${:}\longrightarrow$$a={\frac{6.6}{9}}$

${:}\longrightarrow$$a=0.73m/s {}^{2}$

• According to first equation of motion

${:}\longrightarrow$${\boxed{v=u+at}}$

[by putting values]

${:}\longrightarrow$$3=0+0.73×t$

${:}\longrightarrow$$3=0.73t$

${:}\longrightarrow$$t={\frac {3}{0.73}}$

${:}\longrightarrow$${\underline{\boxed{\bf {t=4.1s}}}}$

• According to second equation of motion

${:}\longrightarrow$${\boxed{s=ut+{\frac{1}{2}}at {}^{2}}}$

${:}\longrightarrow$$s=0×4+{\frac{1 }{2}}×0.73×{4}^{2}$

${:}\longrightarrow$$s={\frac {0.73×4×4}{2}}$

${:}\longrightarrow$$0.73×8$

${:}\longrightarrow$${\underline{\boxed{\bf {S=5.8m}}}}$