It depends on how you want to solve it you can solve it in many different meathods:$
Answer:
Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
![W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5E%7Bx_f%7D_%7Bx_0%7D%20%7BF%28x%29%7D%20%5C%2C%20dx%20%5C%5CW%3D%5Cint%5Climits%5E%7B14%7D_%7B0%7D%20%28%7B18N-0.530%5Cfrac%7BN%7D%7Bm%7Dx%7D%29%20%5C%2C%20dx%5C%5CW%3D%5B%2818N%29x-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7Bx%5E2%7D%7B2%7D%5D%5E%7B14%7D_%7B0%7D%5C%5CW%3D%2818N%2914m-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B%2814m%29%5E2%7D%7B2%7D-%2818N%290%2B%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B0%5E2%7D%7B2%7D%5C%5CW%3D252N%5Ccdot%20m-52N%5Ccdot%20m%5C%5CW%3D200N%5Ccdot%20m)
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:
The box is initially at rest, so 
. Solving for 
:
Answer:
F = 4.47 10⁻⁶ N
Explanation:
The expression they give for the strength of the tide is
F = 2 G m M a / r³
Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg
They ask us to perform the calculation
F = 2 6.67 10⁻¹¹ 135 5.98 10²⁴ 13 / (6.79 10⁶)³
F = 4.47 10⁻⁶ N
This force is directed in the single line at the astronaut's mass centers and the space station
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
Answer:
<h3>2,321.62Joules</h3>
Explanation:
The formula for calculating workdone is expressed as;
Workdone = Force * Distance
Get the force
F = nR
n is the coefficient of friction = 0.5
R is the reaction = mg
R = 46 ( 9.8)
R = 450.8N
F = 0.5 * 450.8
F = 225.4N
Distance = 10.3m
Get the workdone
Workdone = 225.4 * 10.3
Workdone = 2,321.62Joules
<em>Hence the amount of work done is 2,321.62Joules</em>
