A block of mass 1 kg,moving along the x-axis | vedclass

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

Question

(D) Given: mass $m = 1 \ kg$,initial velocity $v_{i} = 10 \ m/s$,force $F = -kx$,where $k = 10 \ N/m$. Using Newton's second law,$a = \frac{F}{m} = -\

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given: mass $m = 1 \ kg$,initial velocity $v_{i} = 10 \ m/s$,force $F = -kx$,where $k = 10 \ N/m$. Using Newton's second law,$a = \frac{F}{m} = -\frac{kx}{m} = -\frac{10x}{1} = -10x$. We know that $a = v \frac{dv}{dx}$,so $v \frac{dv}{dx} = -10x$. Integrating both sides: $\int_{10}^{v} v \, dv = \int_{0.1}^{1.9} -10x \, dx$. $\left[ \frac{v^2}{2} \right]_{10}^{v} = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9}$. $\frac{v^2 - 100}{2} = -5 \left( 1.9^2 - 0.1^2 \right)$. $\frac{v^2 - 100}{2} = -5 \left( 3.61 - 0.01 \right) = -5 \left( 3.60 \right) = -18$. $v^2 - 100 = -36$. $v^2 = 64$. $v = 8 \ m/s$.

Similar Questions

$A$ force $F = 20 + 10y$ acts on a particle in the $y$ direction,where $F$ is in newton $(N)$ and $y$ is in meter $(m)$. The work done by this force to move the particle from $y = 0$ to $y = 1 \; m$ is ...... $J$.

$A$ position-dependent force $F = 7 - 2x + 3x^2 \, N$ acts on a small body of mass $2 \, kg$ and displaces it from $x = 0$ to $x = 5 \, m$. The work done in joules is

$A$ particle of mass $500 \text{ g}$ is at rest. It is free to move along a straight line. The power delivered to the particle varies with time according to the following graph. The momentum of the particle at $t = 5 \text{ s}$ is

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 \, m$ is (in $ \, J$)

The figure shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s = 0$ to $20\, m$ will be......$J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module