Answer:
<h3>The answer is 19,800 J</h3>
Explanation:
The work done by an object can be found by using the formula
<h3>workdone = force × distance</h3>
From the question
force = 120 N
distance = 165 m
We have
work done = 120 × 165
We have the final answer as
<h3>19,800 J</h3>
Hope this helps you
Answer:
<h2>
206.67N</h2>
Explanation:
The sum of force along both components x and y is expressed as;
![\sum Fx = ma_x \ and \ \sum Fy = ma_y](https://tex.z-dn.net/?f=%5Csum%20Fx%20%3D%20ma_x%20%20%5C%20and%20%5C%20%5Csum%20Fy%20%3D%20ma_y)
The magnitude of the net force which is also known as the resultant will be expressed as ![R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}](https://tex.z-dn.net/?f=R%20%3D%5Csqrt%7B%28%5Csum%20Fx%29%5E2%20%2B%20%28%5Csum%20Fx%20%29%5E2%7D)
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
![a_x = \frac{d^2 x }{dt^2}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7Bd%5E2%20x%20%7D%7Bdt%5E2%7D)
![a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28%5Cfrac%7Bdx%7D%7Bdt%7D%20%29%5C%5C%20%5C%5Ca_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%286t%5E%7B2%7D-4%20%20%29%5C%5C%5C%5Ca_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%2812t%20%20%29%5C%5C%5C%5Ca_x%20%3D%2012m%2Fs%5E%7B2%7D)
Similarly,
![a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2](https://tex.z-dn.net/?f=a_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28%5Cfrac%7Bdy%7D%7Bdt%7D%20%29%5C%5C%20%5C%5Ca_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%285t%5E%7B3%7D%20%2B6%20%29%5C%5C%5C%5Ca_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%2815t%5E%7B2%7D%20%20%20%29%5C%5C%5C%5Ca_y%20%3D%2030t%5C%5Ca_y%20%5C%20at%20%5C%20t%3D%202.15s%3B%20a_y%20%3D%2030%282.15%29%5C%5Ca_y%20%3D%2064.5m%2Fs%5E2)
![\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N](https://tex.z-dn.net/?f=%5Csum%20F_x%20%3D%203.15%20%2A%2012%20%3D%2037.8N%5C%5C%5Csum%20F_y%20%3D%203.15%20%2A%2064.5%20%3D%20203.18N)
![R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7B37.8%5E2%2B203.18%5E2%7D%5C%5C%20%5C%5CR%20%3D%20%5Csqrt%7B1428.84%2B41%2C282.11%7D%5C%5C%20%5C%5CR%20%3D%20%5Csqrt%7B42.710.95%7D%5C%5C%20%5C%5CR%20%3D%20206.67N)
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
speed equals distance times time
The magnitude of the average force that the ball exerts against his glove is 600 N
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Further explanation</h3>
Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
![\boxed {F = ma }](https://tex.z-dn.net/?f=%5Cboxed%20%7BF%20%3D%20ma%20%7D)
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<u>Given:</u>
mass of ball = m = 0.15 kg
initial speed of ball = u = 40 m/s
final speed of ball = v = 0 m/s
distance = d = 20 cm = 0.2 m
<u>Asked:</u>
average force = F = ?
<u>Solution:</u>
<em>We will use </em><em>Newton's Law of Motion</em><em> to solve this problem as follows:</em>
![F = m a](https://tex.z-dn.net/?f=F%20%3D%20m%20a)
![F = m (\frac { u^2 - v^2 } { 2d } )](https://tex.z-dn.net/?f=F%20%3D%20m%20%28%5Cfrac%20%7B%20u%5E2%20-%20v%5E2%20%7D%20%7B%202d%20%7D%20%29)
![F = 0.15 \times \frac { 40^2 - 0^2 } { 2 \times 0.2 }](https://tex.z-dn.net/?f=F%20%3D%200.15%20%5Ctimes%20%5Cfrac%20%7B%2040%5E2%20-%200%5E2%20%7D%20%7B%202%20%5Ctimes%200.2%20%7D)
![F = 0.15 \times \frac { 1600 } { 0.4 }](https://tex.z-dn.net/?f=F%20%3D%200.15%20%5Ctimes%20%5Cfrac%20%7B%201600%20%7D%20%7B%200.4%20%7D)
![F = 0.15 \times 4000](https://tex.z-dn.net/?f=F%20%3D%200.15%20%5Ctimes%204000)
![\boxed {F = 600 \texttt{ N}}](https://tex.z-dn.net/?f=%5Cboxed%20%7BF%20%3D%20600%20%5Ctexttt%7B%20N%7D%7D)
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Learn more</h3>
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Dynamics