A force F pulls a block of mass 60 kg at a c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: mass $m = 60 \ kg$,angle $	heta = 37^{\circ}$,coefficient of friction $\mu = 1/3$,$g = 10 \ m/s^2$.Since the block moves at a constant spe

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) Given: mass $m = 60 \ kg$,angle $	heta = 37^{\circ}$,coefficient of friction $\mu = 1/3$,$g = 10 \ m/s^2$.Since the block moves at a constant speed,the net force is zero $(a = 0)$.Resolving the force $F$ into horizontal and vertical components:Horizontal direction: $F \cos 37^{\circ} = \mu N$  --- $(1)$Vertical direction: $N + F \sin 37^{\circ} = mg$  --- $(2)$From $(2)$,$N = mg - F \sin 37^{\circ}$.Substitute $N$ into $(1)$:$F \cos 37^{\circ} = \mu (mg - F \sin 37^{\circ})$$F \cos 37^{\circ} = \mu mg - \mu F \sin 37^{\circ}$$F (\cos 37^{\circ} + \mu \sin 37^{\circ}) = \mu mg$$F = \frac{\mu mg}{\cos 37^{\circ} + \mu \sin 37^{\circ}}$Using $\cos 37^{\circ} = 4/5$ and $\sin 37^{\circ} = 3/5$:$F = \frac{(1/3) 	imes 60 	imes 10}{(4/5) + (1/3) 	imes (3/5)} = \frac{200}{(4/5) + (1/5)} = \frac{200}{1} = 200 \ N$.

$A$ force $F$ pulls a block of mass $60 \ kg$ at a constant speed across the floor. The coefficient of friction is $\mu = 1/3$. What is the magnitude of the force $F$ (in $N$)?

Similar Questions

When a body is moving on a surface,the force of friction is called

$A$ body is travelling with $10 \,ms^{-1}$ on a rough horizontal surface. Its velocity after $2 \,s$ is $4 \,ms^{-1}$. The coefficient of kinetic friction between the block and the plane is (acceleration due to gravity $= 10 \,ms^{-2}$)

The retarding acceleration of $7.35\, ms^{-2}$ due to frictional force stops a car of mass $400\, kg$ travelling on a road. The coefficient of friction between the tyre of the car and the road is

When we rub our palms,they get heated,but only up to a maximum temperature because:

Vedclass Test Series

Exam Paper Generator

Online Exam Module