Calculate the Length & Potential Energy of a Sliding Box on a Ramp

[Shah 72]

A ramp rises 10cm for every 80cm along the sloping surface. A box of mass 50 kg slides down the ramp starting from rest at the top of the ramp. The coefficient of friction between the ramp and the box is 0.03 and no other resistance acts.
The box is traveling 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.

 

[skeeter]

$\theta$ = ramp angle
$\Delta x$ = length of the ramp

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

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

 

[Shah 72]

$\theta$ = ramp angle
$\Delta x$ = length of the ramp

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

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

Thank you!

 

[Shah 72]

$\theta$ = ramp angle
$\Delta x$ = length of the ramp

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

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

$\theta$ = ramp angle
$\Delta x$ = 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.
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.

 

[skeeter]

Why are you messing around with Newton’s 2nd law? … solve the energy equation I posted for $\Delta x$, then substitute in your given values.

 

[Shah 72]

Why are you messing around with Newton’s 2nd law? … solve the energy equation I posted for $\Delta x$, then substitute in your given values.

I haven't yet learned this method.
So I get distance = 100m
But the textbook ans says 2.10m

 

[DaalChawal]

I haven't yet learned this method.

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.

 

[skeeter]

I haven't yet learned this method.
So I get distance = 100m
But the textbook ans says 2.10m

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 :poop:

 

[Shah 72]

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 :poop:

Thanks a lotttt!

 

[Shah 72]

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.

Thanks!