A particle is acted upon by a force F which v | vedclass

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

Question

(A) The work done by a variable force is equal to the area under the force-displacement $(F-x)$ graph.Area from $x = 0$ to $x = 6$: $	ext{Area}_1 = \

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) The work done by a variable force is equal to the area under the force-displacement $(F-x)$ graph.Area from $x = 0$ to $x = 6$: $	ext{Area}_1 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 6 	imes 10 = 30\,J$.Area from $x = 6$ to $x = 9$: $	ext{Area}_2 = 	ext{length} 	imes 	ext{width} = (9 - 6) 	imes (-5) = 3 	imes (-5) = -15\,J$.Area from $x = 9$ to $x = 13$: $	ext{Area}_3 = 	ext{length} 	imes 	ext{width} = (13 - 9) 	imes 5 = 4 	imes 5 = 20\,J$.Area from $x = 13$ to $x = 16$: $	ext{Area}_4 = 	ext{length} 	imes 	ext{width} = (16 - 13) 	imes (-5) = 3 	imes (-5) = -15\,J$.Total work done $W = 	ext{Area}_1 + 	ext{Area}_2 + 	ext{Area}_3 + 	ext{Area}_4 = 30 - 15 + 20 - 15 = 20\,J$.According to the work-energy theorem,$W = K_f - K_i$.Given $K_i = 25\,J$,we have $20 = K_f - 25$.Therefore,$K_f = 20 + 25 = 45\,J$.

$A$ particle is acted upon by a force $F$ which varies with position $x$ as shown in the figure. If the particle at $x = 0$ has a kinetic energy of $25\,J$,then the kinetic energy of the particle at $x = 16\,m$ is .............. $J$.

Similar Questions

The slope of the kinetic energy-displacement curve of a particle in motion is

The acceleration-displacement graph of a $10 \ kg$ particle is given. Find the work done as the particle moves from $x = 0 \ cm$ to $x = 8 \ cm$.

Power applied to a particle varies with time as $P = (3t^2 - 2t + 1) \text{ W}$. The change in kinetic energy of the particle from $t = 2 \text{ s}$ to $t = 4 \text{ s}$ is ............... $J$.

$A$ force $\vec{F} = (-6x^3 \hat{i}) \ N$ acts on a particle,causing a displacement from $x = 4 \ m$ to $x = -2 \ m$. The work done by the force is .......... $J$.

$A$ force acts on a particle moving along the $x$-axis,and its variation with position is shown in the figure. At which point will the particle be in a stable equilibrium position?

Vedclass Test Series

Exam Paper Generator

Online Exam Module