Math Problem Statement

A 8.90-kg block of ice, released from rest at the top of a 1.32-m -long frictionless ramp, slides downhill, reaching a speed of 2.99 m/s at the bottom. What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.1 N parallel to the surface of the ramp?

Solution

To solve this problem, we will use principles of energy conservation and work-energy theorem.

Step 1: Work without friction

First, let's calculate the energy and speed of the block when there is no friction (already provided):

  • Mass of the block m=8.90kgm = 8.90 \, \text{kg}
  • Length of the ramp d=1.32md = 1.32 \, \text{m}
  • Initial speed vi=0m/sv_i = 0 \, \text{m/s}
  • Final speed without friction vf=2.99m/sv_f = 2.99 \, \text{m/s}

The mechanical energy is conserved. The initial potential energy (PE) of the block is completely converted into kinetic energy (KE) at the bottom:

PE=KEPE = KE

mgh=12mvf2mgh = \frac{1}{2}mv_f^2

where hh is the vertical height of the ramp.

We can find the height using the length of the ramp and the fact that the ramp is inclined:

h=vf22gh = \frac{v_f^2}{2g}

Step 2: With Friction

Now, when friction is acting, the work-energy theorem states:

KEfinal=PEinitialWfrictionKE_{final} = PE_{initial} - W_{friction} where:

  • Wfriction=fdW_{friction} = f \cdot d, and f=10.1Nf = 10.1 \, \text{N},

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Energy Conservation
Work-Energy Theorem
Friction

Formulas

Potential Energy (PE) = mgh
Kinetic Energy (KE) = (1/2)mv^2
Work Done by Friction = Force × Distance
Work-Energy Theorem: KE_final = PE_initial - W_friction

Theorems

Work-Energy Theorem

Suitable Grade Level

Grades 11-12