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

• MHB
• Shah 72
In summary: 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$Shah 72 MHB A ramp rises 10cm for every 80cm along the sloping surface. A box of mass 50 kg slides down the ramp starting from rest at the top of the ramp. The coefficient of friction between the ramp and the box is 0.03 and no other resistance acts. The box is traveling at 2 m/s when it reaches the bottom of the ramp. Find the length of the ramp Find the loss in the potential energy of the box. I don't understand how to calculate. Pls help.$\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$skeeter said:$\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$Thank you! skeeter said:$\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$skeeter said:$\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$I tried doing but not getting the ans. m = 50 kg, coefficient of friction is 0.03 F= m×a 500sin theta- 0.03 × 500 cos theta= 50a I don't know how to calculate the length of the ramp. Pls can you help. Why are you messing around with Newton’s 2nd law? … solve the energy equation I posted for$\Delta x$, then substitute in your given values. skeeter said: Why are you messing around with Newton’s 2nd law? … solve the energy equation I posted for$\Delta x$, then substitute in your given values. I haven't yet learned this method. So I get distance = 100m But the textbook ans says 2.10m Shah 72 said: I haven't yet learned this method. This is just derived from F= ma only$F= m v \frac{dv}{ds} F ds = m v dv $integrate this you will get the result also known as Work energy theorem. Shah 72 said: I haven't yet learned this method. So I get distance = 100m But the textbook ans says 2.10m well, it's about time you learned ... let the bottom of the ramp be the zero for gravitational potential energy (initial quantity of mechanical energy) - (work done by friction) = (final quantity of mechanical energy) the block starts at rest, so the initial quantity of mechanical energy is gravitational potential energy,$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$$mg \Delta x \sin{\theta} - \mu mg \cos{\theta} \Delta x = \dfrac{1}{2}mv_f^2 \Delta x(g \sin{\theta} - \mu g \cos{\theta}) = \dfrac{1}{2}v_f^2 \color{red} \Delta x = \dfrac{v_f^2}{2g(\sin{\theta} - \mu\cos{\theta})} I get \Delta x = 2.1 \, mLet's check by messing with Newton's 2nd ... a = \dfrac{F_{net}}{m} = \dfrac{mg \sin{\theta} - \mu mg \cos{\theta}}{m} a = g(\sin{\theta} - \mu \cos{\theta}) v_f^2 = v_0^2 + 2a \Delta x \implies \Delta x = \dfrac{v_f^2 - v_0^2}{2a}  \color{red} \Delta x = \dfrac{v_f^2}{2g(\sin{\theta} - \mu \cos{\theta})} Well ... I'll be dipped in dog :poop: skeeter said: well, it's about time you learned ... let the bottom of the ramp be the zero for gravitational potential energy (initial quantity of mechanical energy) - (work done by friction) = (final quantity of mechanical energy) the block starts at rest, so the initial quantity of mechanical energy is gravitational potential energy, mgh_0 where h_0 = \Delta x \sin{\theta} work done by kinetic energy is f \cdot \Delta x = \mu mg\cos{\theta} \Delta x final quantity of mechanical energy is strictly kinetic, K = \dfrac{1}{2}mv_f^2$$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})}$Well ... I'll be dipped in dog :poop: Thanks a lotttt! DaalChawal said: This is just derived from F= ma only$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!

## 1. How do you calculate the length of a sliding box on a ramp?

To calculate the length of a sliding box on a ramp, you will need to know the height of the ramp, the angle of the ramp, and the horizontal distance traveled by the box. You can then use the formula L = H/sin(θ) to calculate the length, where L is the length, H is the height, and θ is the angle of the ramp in radians.

## 2. What is potential energy and how is it related to a sliding box on a ramp?

Potential energy is the energy an object has due to its position or state. In the case of a sliding box on a ramp, potential energy is the energy the box has due to its position on the ramp. As the box slides down the ramp, its potential energy decreases and is converted into kinetic energy.

## 3. How do you calculate the potential energy of a sliding box on a ramp?

The potential energy of a sliding box on a ramp can be calculated using the formula PE = mgh, where PE is the potential energy, m is the mass of the box, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the ramp.

## 4. What factors can affect the potential energy of a sliding box on a ramp?

The potential energy of a sliding box on a ramp can be affected by the mass of the box, the height and angle of the ramp, and the acceleration due to gravity. Additionally, friction and air resistance can also affect the potential energy of the box.

## 5. How can the potential energy of a sliding box on a ramp be used in real-life applications?

The potential energy of a sliding box on a ramp can be used in real-life applications such as roller coasters and water slides. In these cases, the potential energy of the riders at the top of the ride is converted into kinetic energy as they slide down the ramp, providing an exciting and thrilling experience.

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