Physics, asked by wwwnivasgeetha303, 5 hours ago

a force of 2 n acts on a body of mass 2.0 kg .as a result body covers a distance 1 m in 4 seconds moving along a straight line find the intial velocity

Answers

Answered by Anonymous
1

Answer:

Answer

Given:

The mass of the body is m=2kg

The force applied on the body F=7N

The coefficient of kinetic friction =0.1

Since the body starts from rest, the initial velocity of body is zero.

Time at which the work is to be determined is t=10s

The acceleration produced in the body by the applied force is given by Newtons second law of motion as:

a

=

m

F

=

2

7

=3.5m/s

2

Frictional force is given as:

f=μg=0.1×2×9.8=1.96

The acceleration produced by the frictional force:

a"=−

2

1.96

=−0.98m/s

2

Therefore, the total acceleration of the body:

a

+a"=3.5+(−0.98)=2.52m/s

2

The distance traveled by the body is given by the equation of motion:

s=ut+

2

1

at

2

=0+

2

1

×2.52×(10)

2

=126 m

(a) Work done by the applied force,

W

a

=F⋅s=7×126=882 J

(b) Work done by the frictional force,

W

f

=F⋅s=1.96×126=247 J

(c), (d)

From the first equation of motion, final velocity can be calculated as:

v=u+at

=0+2.52×10=25.2m/s

So, the change in kinetic energy is

ΔK=

2

1

mv

2

2

1

mu

2

=

2

1

2(v

2

−u

2

)=(25.2)

2

−0

2

=635 J

The distance traveled by the body is given by the equation of motion:

s=ut+

2

1

at

2

=0+

2

1

×2.52×(10)

2

=126 m

Net force =7+(1.96)=5.04 N

Work done by the net force,

W

net

=5.04×126=635 J

Explanation:

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Answered by SKASHISH666
1

Explanation:

The mass of the body is m=2kg

The force applied on the body F=7N

The coefficient of kinetic friction =0.1

Since the body starts from rest, the initial velocity of body is zero.

Time at which the work is to be determined is t=10s

The acceleration produced in the body by the applied force is given by Newtons second law of motion as

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