A trolley going down an inclined plane has an acceleration of 2cm/s^2 What will be its velocity

Physics

1 answer:

Snowcat [4.5K]3 years ago

5 0

Answer:

$V_{f}$ = 6 cm/s

Explanation:

Given:

Acceleration = a = 2 cm/s²

Time = t = 3s

Initial Velocity = $V_{i}$ = 0 cm/s

Required:

Velocity = $V_{f}$ = ?

Formula:

a = $\frac{V_{f}-V_{i}}{t}$

Solution:

2 = $\frac{V_{f}-0}{3}$

=> $V_{f}$ = 2*3

=> $V_{f}$ = 6 cm/s

You might be interested in

A 0.20-kg particle moves along the x axis under the influence of a conservative force. The potential energy is given by

anastassius [24]

Answer:

e) 11 m/s

Explanation:

For a particle under the action of a conservative force, its mechanical energy at point 0 is must be equal to its mechanical energy at point 1:

$K_1+U_1=K_0+U_0\\\\\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(x_1)^2+(2.0\frac{J}{m^4})(x_1)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})(x_0)^2+(2.0\frac{J}{m^4})(x_0)^4]$

In $x_1=1.0m$ the speed is given, so $v_1=5.0\frac{m}{s}$ and $x_0=0$ . Replacing:

$\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(1m)^2+(2.0\frac{J}{m^4})(1m)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})0^2+(2.0\frac{J}{m^4})(0)^4]\\\frac{mv_1^2}{2}+8.0J+2.0J=\frac{mv_0^2}{2}\\\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+10J\\v_0=\sqrt{\frac{2}{m}(\frac{mv_1^2}{2}+10J)}\\v_0=\sqrt{\frac{2}{0.2kg}(\frac{(0.2kg)(5\frac{m}{s})^2}{2}+10J)}\\v_0=11.18\frac{m}{s}$