A horizontal force of 10 N is applied on a b | vedclass

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

Question

(C) Given: Force $F = 10 \ N$,mass $m = 1.5 \ kg$,coefficient of kinetic friction $\mu_k = 0.2$,$g = 10 \ m/s^2$,initial velocity $u = 0$,time $t = 6

Vedclass · Accepted Answer

$840$

Vedclass · Answer

(C) Given: Force $F = 10 \ N$,mass $m = 1.5 \ kg$,coefficient of kinetic friction $\mu_k = 0.2$,$g = 10 \ m/s^2$,initial velocity $u = 0$,time $t = 6 \ s$.First,calculate the kinetic friction force: $f_k = \mu_k \cdot m \cdot g = 0.2 	imes 1.5 	imes 10 = 3 \ N$.The net force acting on the block is $F_{net} = F - f_k = 10 - 3 = 7 \ N$.The acceleration of the block is $a = F_{net} / m = 7 / 1.5 = 14 / 3 \ m/s^2$.The displacement $s$ in $t = 6 \ s$ is given by $s = ut + (1/2)at^2 = 0 + (1/2) 	imes (14/3) 	imes (6)^2 = (1/2) 	imes (14/3) 	imes 36 = 7 	imes 12 = 84 \ m$.The work done by the applied force is $W = F 	imes s = 10 	imes 84 = 840 \ J$.

Similar Questions

$A$ string is used to pull a block of mass $m$ vertically up by a distance $h$ at a constant acceleration $\frac{g}{4}$. The work done by the tension in the string is ...............

$A$ block of mass $M$ is pulled a distance $x$ on a horizontal table. The work done by the weight (gravitational force) is:

When an object is dragged $10 \, m$ along a road,a frictional force of $200 \, N$ acts in the direction opposite to its motion. What is the work done by the frictional force on the object (in $, J$)?

If a force $\vec{F}=(3 \hat{i}-2 \hat{j}) \text{ N}$ acting on a body displaces it from point $(1 \text{ m}, 2 \text{ m})$ to point $(2 \text{ m}, 0 \text{ m})$,then the work done by the force is (in $\text{ J}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module