Question

a mass 5 kg is acted upon by a force of 1 n starting from rest how much is distance covered by the mass in 20s?​

Answer 1

Given :-

• A mass 5 kg is acted upon by a force of 1 n starting from rest

To find :-

• how much is distance covered by the body in 20s?

Solution :-

• Mass of body = 5 kg

• Force acted = 1 N

As we know that

→ Force = Mass × acceleration

→ F = m × a

→ 1 = 5 × a

→ a = 1/5 = 0.2 m/s²

Now,

• Acceleration = 0.2 m/s²

• Initial velocity = 0

• Time = 20s

According to second equation of motion

→ s = ut + ½ at²

Where " s " is distance covered, " u " is initial velocity, " a " is acceleration and " t " is time.

• According to the question

→ s = ut + ½ at²

→ s = 0 × 20 + ½ × 0.2 × (20)²

→ s = 0 + 0.1 × 400

→ s = 40 m

Hence,

• Distance covered i.e 40 m

Answer 2

Explanation:

Given:-

Mass=m=5kg

Force=F=1N

Time=t=20s

Intial velocity=u=0

To find:-

Distance covered=s

Solution:-

As we know that

${:}\longrightarrow$${\boxed {Force=Mass×Acceleration}}$

${:}\longrightarrow$$F=m×a$

${:}\longrightarrow$$1=5×a$

${:}\longrightarrow$$a={\frac{1}{5}}$

${:}\longrightarrow$$a=0.2m/s {}^{2}$

According to the second equation of motion:-

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

Where "s"=distance covered,"u"=intial velocity,"a"=acceleration,"t"=time.

According to the question :-

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

${:}\longrightarrow$$s=0×20+{\frac {1}{2}}×0.2×20 {}^{2}$

${:}\longrightarrow$$s=0+0.1×400$

${:}\longrightarrow$${\underline{\boxed{\bf{s=40m}}}}$

Hence,

Distance covered=40m