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Math  /  Algebra

Question38. Suppose that the particle of the previous problem also experiences forces F⃗2=−15i^ N\vec{F}_{2}=-15 \hat{i} \mathrm{~N} and F⃗3=6.0j^ N\vec{F}_{3}=6.0 \hat{j} \mathrm{~N}. What is its acceleration in this
39. Find the acceleration of the body of mass 5.0 kg shown below.

Studdy Solution

STEP 1

1. The particle is subject to multiple forces, and we need to find the net force acting on it.
2. The mass of the particle is given as 5.0 kg.
3. Newton's second law of motion, F⃗=ma⃗ \vec{F} = m \vec{a} , applies, where F⃗ \vec{F} is the net force, m m is the mass, and a⃗ \vec{a} is the acceleration.
4. Forces are given in vector form, and we need to consider both the x and y components.

STEP 2

1. Determine the net force acting on the particle.
2. Calculate the acceleration using Newton's second law.

STEP 3

First, identify all the forces acting on the particle and express them in vector form. The forces given are:
- F⃗1=10.0i^ N \vec{F}_1 = 10.0 \hat{i} \, \mathrm{N} (force in the positive x-direction) - F⃗2=−15i^ N \vec{F}_2 = -15 \hat{i} \, \mathrm{N} (force in the negative x-direction) - F⃗3=6.0j^ N \vec{F}_3 = 6.0 \hat{j} \, \mathrm{N} (force in the positive y-direction) - F⃗4=25.0j^ N \vec{F}_4 = 25.0 \hat{j} \, \mathrm{N} (force in the positive y-direction)

STEP 4

Calculate the net force in the x-direction by summing the x-components of all forces:
Fnet,x=10.0 N+(−15 N)=−5.0 N F_{\text{net}, x} = 10.0 \, \mathrm{N} + (-15 \, \mathrm{N}) = -5.0 \, \mathrm{N}

STEP 5

Calculate the net force in the y-direction by summing the y-components of all forces:
Fnet,y=6.0 N+25.0 N=31.0 N F_{\text{net}, y} = 6.0 \, \mathrm{N} + 25.0 \, \mathrm{N} = 31.0 \, \mathrm{N}

STEP 6

Combine the net forces in the x and y directions to find the total net force vector:
F⃗net=Fnet,xi^+Fnet,yj^=−5.0i^+31.0j^ N \vec{F}_{\text{net}} = F_{\text{net}, x} \hat{i} + F_{\text{net}, y} \hat{j} = -5.0 \hat{i} + 31.0 \hat{j} \, \mathrm{N}

STEP 7

Use Newton's second law to find the acceleration vector. Given the mass m=5.0 kg m = 5.0 \, \mathrm{kg} , the acceleration is:
a⃗=F⃗netm=−5.0i^+31.0j^5.0 \vec{a} = \frac{\vec{F}_{\text{net}}}{m} = \frac{-5.0 \hat{i} + 31.0 \hat{j}}{5.0}
a⃗=−1.0i^+6.2j^ m/s2 \vec{a} = -1.0 \hat{i} + 6.2 \hat{j} \, \mathrm{m/s^2}
The acceleration of the particle is:
−1.0i^+6.2j^ m/s2 \boxed{-1.0 \hat{i} + 6.2 \hat{j} \, \mathrm{m/s^2}}

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