Question

A block of mass 2.5 Kg is pushed 2.20 m along a frictionless horizontal table by a constant 16N
directed 45° above the horizontal. The work done by the normal force exerted by the table is
a) 24.90
b) 29.4, c) zero
d) 942
​

Answer 1

❏ Formula Used:-

✦ If F be the force applied on a body of mass m, along the displacement of the body (or along horizontal plane) and if s be the displacement, then.

➝The work done (W) on the body is given by,

$\sf\longrightarrow\boxed{ W=F\times s}$

✦if f is applied with an angle $\bf\theta$with the horizontal plane ,

➝Then the work done (W) is given by,

$\sf\longrightarrow \boxed{W=(F\cos\theta\times s)=Fs\cos\theta}$

❏ Question:-

@A block of mass 2.5 Kg is pushed 2.20 m along a frictionless horizontal table by a constant 16N directed 45° above the horizontal.

The work done by the normal force exerted by the table is,

a) 24.90 Joule

b) 29.4, Joule

c) O Joule

d) 942 Joule

❏ Solution:-

❚ Given:-

• $\bf\theta_f=45\degree$

• mass(m)=2.5 kg

• distance(s)=2.20 m

• Force(F)= 16 N

❚ To Find:-

• Total work done($\bf W_t$)= ?

❚ Ans:-

➔Work done due to the Applied Force,

• $\sf\longrightarrow W_f=Fs\cos\theta_f$

• $\sf\longrightarrow W_f=(16\times 2.20\times \cos45\degree)\:\:Joule$

• $\sf\longrightarrow W_f=(16\times 2.20\times \frac{1}{\sqrt{2}})\:\:Joule$

• $\sf\longrightarrow \boxed{W_f=24.90\:\:Joule}$

➔Work done due to the Normal force:-

• $\sf\longrightarrow W_n=F_n s\cos\theta_n$

• $\sf\longrightarrow W_n=(mg)\times \cos90\degree)\:\:Joule$

• $\sf\longrightarrow W_n=(mg)\times0\:\:Joule$

• $\sf\longrightarrow \boxed{W_n=0\:\:Joule}$

➔Work done due to the weight ( force):-

• $\sf\longrightarrow W_{mg}=F_n s\cos\theta_{mg}$

• $\sf\longrightarrow W_{mg}=(mg)\times \cos90\degree)\:\:Joule$

• $\sf\longrightarrow W_{mg}=(mg)\times0\:\:Joule$

• $\sf\longrightarrow \boxed{W_{mg}=0\:\:Joule}$

➔Total work done:-

• $\sf\longrightarrow W_{t}=(W_f+W_n+W_{mg})\:\:Joule$

• $\sf\longrightarrow W_{t}=(24.90+0+0)\:\:Joule$

• $\sf\longrightarrow\boxed{ \large{\red{W_{t}=24.90\:\:Joule}}}$

∴ Total work done is =24.90 Joule.

❚∴ Option $\sf\longrightarrow$=(a) 24.90 J.

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Answer 2

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