- Thread starter
- #1

The box is travelling at 2 m/s when it reaches the bottom of the ramp.

Find the length of the ramp

Find the loss in the potential energy of the box.

I don't understand how to calculate. Pls help.

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- Start date

- Thread starter
- #1

The box is travelling at 2 m/s when it reaches the bottom of the ramp.

Find the length of the ramp

Find the loss in the potential energy of the box.

I don't understand how to calculate. Pls help.

- Mar 1, 2012

- 973

$\Delta x$ = length of the ramp

you are given info that can be used to determine $\sin{\theta}$

$mg \cdot \Delta x \sin{\theta} - \mu mg\cos{\theta} \cdot \Delta x = \dfrac{1}{2}m v_f^2$

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- #3

Thank you!

$\Delta x$ = length of the ramp

you are given info that can be used to determine $\sin{\theta}$

$mg \cdot \Delta x \sin{\theta} - \mu mg\cos{\theta} \cdot \Delta x = \dfrac{1}{2}m v_f^2$

- Thread starter
- #4

$\Delta x$ = length of the ramp

you are given info that can be used to determine $\sin{\theta}$

$mg \cdot \Delta x \sin{\theta} - \mu mg\cos{\theta} \cdot \Delta x = \dfrac{1}{2}m v_f^2$

I tried doing but not getting the ans.

$\Delta x$ = length of the ramp

you are given info that can be used to determine $\sin{\theta}$

$mg \cdot \Delta x \sin{\theta} - \mu mg\cos{\theta} \cdot \Delta x = \dfrac{1}{2}m v_f^2$

m = 50 kg, coefficient of friction is 0.03

F= m×a 500sin theta- 0.03 × 500 cos theta= 50a

I don't know how to calculate the length of the ramp.

Pls can you help.

- Mar 1, 2012

- 973

- Thread starter
- #6

I haven't yet learnt this method.

So I get distance = 100m

But the textbook ans says 2.10m

- Apr 8, 2021

- 64

This is just derived from F= ma onlyI haven't yet learned this method.

$F= m v \frac{dv}{ds}$

$ F ds = m v dv $

integrate this you will get the result also known as Work energy theorem.

- Mar 1, 2012

- 973

well, it's about time you learned ...I haven't yet learnt this method.

So I get distance = 100m

But the textbook ans says 2.10m

let the bottom of the ramp be the zero for gravitational potential energy

(initial quantity of mechanical energy) - (work done by friction) = (final quantity of mechanical energy)

the block starts at rest, so the initial quantity of mechanical energy is gravitational potential energy, $mgh_0$ where $h_0 = \Delta x \sin{\theta}$

work done by kinetic energy is $f \cdot \Delta x = \mu mg\cos{\theta} \Delta x$

final quantity of mechanical energy is strictly kinetic, $K = \dfrac{1}{2}mv_f^2$

$mg \Delta x \sin{\theta} - \mu mg \cos{\theta} \Delta x = \dfrac{1}{2}mv_f^2$

$\Delta x(g \sin{\theta} - \mu g \cos{\theta}) = \dfrac{1}{2}v_f^2$

$\color{red} \Delta x = \dfrac{v_f^2}{2g(\sin{\theta} - \mu\cos{\theta})}$

I get $\Delta x = 2.1 \, m$

Let's check by messing with Newton's 2nd ...

$a = \dfrac{F_{net}}{m} = \dfrac{mg \sin{\theta} - \mu mg \cos{\theta}}{m}$

$a = g(\sin{\theta} - \mu \cos{\theta})$

$v_f^2 = v_0^2 + 2a \Delta x \implies \Delta x = \dfrac{v_f^2 - v_0^2}{2a} $

$\color{red} \Delta x = \dfrac{v_f^2}{2g(\sin{\theta} - \mu \cos{\theta})}$

Well ... I'll be dipped in dog

- Thread starter
- #9

Thanks a lotttt!!!well, it's about time you learned ...

let the bottom of the ramp be the zero for gravitational potential energy

(initial quantity of mechanical energy) - (work done by friction) = (final quantity of mechanical energy)

the block starts at rest, so the initial quantity of mechanical energy is gravitational potential energy, $mgh_0$ where $h_0 = \Delta x \sin{\theta}$

work done by kinetic energy is $f \cdot \Delta x = \mu mg\cos{\theta} \Delta x$

final quantity of mechanical energy is strictly kinetic, $K = \dfrac{1}{2}mv_f^2$

$mg \Delta x \sin{\theta} - \mu mg \cos{\theta} \Delta x = \dfrac{1}{2}mv_f^2$

$\Delta x(g \sin{\theta} - \mu g \cos{\theta}) = \dfrac{1}{2}v_f^2$

$\color{red} \Delta x = \dfrac{v_f^2}{2g(\sin{\theta} - \mu\cos{\theta})}$

I get $\Delta x = 2.1 \, m$

Let's check by messing with Newton's 2nd ...

$a = \dfrac{F_{net}}{m} = \dfrac{mg \sin{\theta} - \mu mg \cos{\theta}}{m}$

$a = g(\sin{\theta} - \mu \cos{\theta})$

$v_f^2 = v_0^2 + 2a \Delta x \implies \Delta x = \dfrac{v_f^2 - v_0^2}{2a} $

$\color{red} \Delta x = \dfrac{v_f^2}{2g(\sin{\theta} - \mu \cos{\theta})}$

Well ... I'll be dipped in dog

- Thread starter
- #10

Thanks!!This is just derived from F= ma only

$F= m v \frac{dv}{ds}$

$ F ds = m v dv $

integrate this you will get the result also known as Work energy theorem.