Question

A boy pulls a $5kg$ block $20meters$ along a horizontal surface at a constant speed with a force directed ${45}^o$ above the horizontal. If the coefficient of kinetic friction is $0.20,$ how much work does the body do on the block?

• A. $163.32J$
• B. $11.55J$
• C. $150J$
• D. $115J$

Answer 1

The correct option is A $163.32J$

Given mass(m)= 5 kg; displacement (s)=20 m;${\mu }_k$=0.20; $\theta$=45

Normal force (vertical component)=F sin${45}^0$

Applied Force =F cos${45}^0$

Net force$F_{net}=mgFsin{45}^{o}C$

Frictional force f=$\mu (mgFsin{45)}^{o}C$

$Fcos{45}^0=\mu \left(mgFsin{45}^0\right)$

$Fcos{45}^0+\mu Fsin{45}^0=\mu mg$

$Fcos{45}^0=\frac{\mu mg}{1+\mu tan{45}^0}$.................(1)

Also we know work done =Force$\times$ displacement

=$\left(\frac{\mu mg}{1+\mu }\right)\times s=\left(\frac{0.2\times 5\times 9.81}{1.2}\right)\times 20=163.5J$