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Chapter 4: Problem 94

A 10.0 -kg block is released from rest on a frictionless track inclined at an angle of \(55^{\circ} .\) (a) What is the net force on the block after it is released? (b) What is the acceleration of the block? (c) If the block is released from rest, how long will it take for the block to attain a speed of \(10.0 \mathrm{m} / \mathrm{s} ?\) (d) Draw a motion diagram for the block. (e) Draw a graph of \(v_{x}(t)\) for values of velocity between 0 and $10 \mathrm{m} / \mathrm{s}\(. Let the positive \)x$ -axis point down the track.

Short Answer

Expert verified
Answer: The net force acting on the block is 80.0 N, and its acceleration is 8.0 m/s². It takes 1.25 seconds for the block to reach a velocity of 10.0 m/s.

Step by step solution

01

Finding the gravitational force acting on the block

First, find the gravitational force acting on the block. This is given by: $$ F_g = m \times g $$ where \(m=10.0\,\text{kg}\) is the mass of the block, and \(g=9.81\,\text{m/s}^2\) is the acceleration due to gravity. Substituting the values, we get: $$ F_g = 10.0\,\text{kg} \times 9.81\,\text{m/s}^2 = 98.1\,\text{N} $$
02

Decomposing the gravitational force into components

We can decompose the gravitational force acting on the block into two components: one parallel and one perpendicular to the inclined surface. Given the angle of inclination \(\theta=55^{\circ}\), $$ F_{gx} = F_g \sin\theta \quad \text{and}\quad F_{gy} = F_g \cos\theta $$ Plugging in the calculated \(F_g=98.1\,\text{N}\) and given angle \(\theta=55^{\circ}\), we get: $$ F_{gx} = 98.1\,\text{N} \sin 55^{\circ} \approx 80.0\,\text{N} $$ $$ F_{gy} = 98.1\,\text{N} \cos 55^{\circ} \approx 56.2\,\text{N} $$
03

Newton's second law of motion

Since there is no external horizontal force acting on the block, using Newton's second law of motion in the x-direction (parallel to the inclined plane) yields: $$ F_{net} = m a_x $$ Using the calculated value for \(F_{gx}=80.0\, \text{N}\), $$ a_x = \frac{F_{gx}}{m} = \frac{80.0\,\text{N}}{10.0\,\text{kg}} = 8.0\,\text{m/s}^2 $$ The net force \(F_{net}\) is equal to \(F_{gx}\), so the net force is \(80.0\,\text{N}\), and the acceleration of the block is \(8.0\,\text{m/s}^2\).
04

Time to reach a velocity of 10.0 m/s

Using the kinematic equation for velocity: $$ v_x = v_{0x} + a_x t $$ where \(v_x\) is the final velocity, \(v_{0x}\) is the initial velocity, \(a_x\) is the acceleration, and \(t\) is the time taken to reach the final velocity. Since the block is released from rest, \(v_{0x} = 0\). We want to find \(t\) for \(v_x=10.0\,\text{m/s}\). $$ 10.0\,\text{m/s} = 0 + 8.0\,\text{m/s}^2 \times t $$ Solving for \(t\), we get: $$ t = \frac{10.0\,\text{m/s}}{8.0\,\text{m/s}^2} = 1.25\,\text{s} $$
05

Motion diagram and v_x(t) graph

(d) In the motion diagram, we can represent the block with a series of arrows. The length of each arrow is proportional to the distance moved and the horizontal distance between arrows represents equal time intervals. The arrows start from rest and gradually become longer, as the block gains speed and moves down the inclined plane. (e) To plot the graph of \(v_x(t)\) for velocities between \(0\,\text{m/s}\) and \(10\,\text{m/s}\), use the equation: $$ v_x(t)=8.0t $$ The graph is a straight line with a slope of \(8.0\,\text{m/s}^2\) and intersects the time axis at \(t=0\) (initially at rest). The velocity increases linearly with respect to time and reaches \(10.0\,\text{m/s}\) at \(t=1.25\,\text{s}\).

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