Question

plz give the answer with proper steps
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Answer 1

Answer:

solved

Explanation:

total mass m = 72 =15 = 87 KG ,   a = 0.6 m/s²

F = ma = 87x0.6 = 52.2 N

a) work = F.d = 52.2x2.5 = 130.5 j

b) d = w/F = 140/52.2 = 2.68 m.

c) of course it depends

because the speed is a function of the acceleration a and the distance d.

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass Of Patient (m₁) = 72 Kg.
• Mass Of Trolley (m₂) = 15 Kg.
• Acceleration (a) = 0.60 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

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Total Mass (M) :-

M = m₁ + m₂

M = 72 + 15

M = 87 Kg.

Applying Newton's second law of motion,

$\large{\boxed{\tt F = Ma}}$

Substituting the values,

$\large{\tt \leadsto F = 87 \times 0.60}$

As,

• Mass = 87 Kg.
• Acceleration = 0.60 m/s².

Therefore,

$\large{\leadsto \underline{\underline{\tt F = 52.2 \: N}}}$

Applying Work Done formula,

$\large{\boxed{\tt W = F.s}}$

Substituting the values,

$\large{\tt \leadsto W = 52.2 \times 2.5}$

As,

• Force we have Found In the above case.
• And The question specifies Displacement as 2.5 meters.

Simplifying,

$\large{\tt \leadsto W = 52.2 \times 2.5}$

$\huge{\boxed{\boxed{\tt W = 130.5 \: J}}}$

So, The work done to move the patient 2.5 meters is 130.5 Joules.

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Now,

The given Work done is 140 Joules And, The applied Force will be same I.e. 52.2 N.

Applying Work done formula,

$\large{\boxed{\tt W = F.s}}$

Substituting the values,

$\large{\tt \leadsto 140 = 52.2 \times s}$

$\large{\tt \leadsto s = \dfrac{140}{52.2}}$

$\large{\tt \leadsto s = 2.68 \: m}$

Taking Approximation,

$\huge{\boxed{\boxed{\tt s = 2.7 \: m}}}$

So, the Person pushes the block 2.7 meters while doing 140 Joules of work.

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Yes, work done depends on The velocity of the body,

For this,

Take Third kinematic equation,

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

Assume the body is at rest,

Therefore

• Initial velocity (u) = 0 m/s.

$\large{\tt \leadsto v^2 - 0 = 2as}$

$\large{\tt \leadsto v^2 = 2as}$

$\large{\tt \leadsto \underline{\underline{s = \dfrac{v^2}{2a}}}}$

Substituting it in the Work done formula,

$\large{\boxed{\tt W = F.s}}$

$\large{\tt \leadsto W = F \times \dfrac{v^2}{2a}}$

$\large{\tt \leadsto W = F . \dfrac{v^2}{2a}}$

$\huge{\boxed{\boxed{\tt W \propto v^2}}}$

Therefore, the Work done by any system depends on the velocity (Directly proportional).

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