Respuesta :
Answer:
Work done on the box is 490 joules.
Explanation:
It is given that,
Mass of the box, m = 5 kg
Initial speed of the box, u = 0
Acceleration of the box, [tex]a=2\ m/s^2[/tex]
Time taken, t = 7 s
The force is calculate using the product of mass and acceleration. It is given by :
F = ma
[tex]F=5\ kg\times 2\ m/s^2=10\ N[/tex]
Let d is the distance covered by the box. Using the equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
[tex]d=\dfrac{1}{2}\times 2\ m/s^2\times (7\ s)^2[/tex]
d = 49 m
The product of force and distance is equal to the work done on the box. It is given by :
[tex]W=F\times d[/tex]
[tex]W=10\ N\times 49\ m[/tex]
W = 490 joules
So, the work done on the box is 490 Joules. Hence, this is the required solution.
The net work done on the box weighing 5.0 kg and accelerated from rest by a force across a floor is 490 Nm.
Given the following data:
- Initial velocity = 0 m/s (since the box starts from rest).
- Mass of box = 5 kg
- Acceleration = 2 [tex]m/s^2[/tex]
- Time = 7 seconds
To find the net work done on the box:
First of all, we would determine the force acting on the box:
[tex]Force = Mass[/tex] × [tex]acceleration[/tex]
[tex]Force = 5[/tex] × [tex]2[/tex]
Force = 10 Newton
Next, we would use the second equation of motion to determine the distance traveled by the box:
[tex]S = ut+ \frac{1}{2}at^2[/tex]
Where:
- S is the displacement or distance traveled.
- u is the initial velocity.
- a is the acceleration.
- t is the time measured in seconds.
Substituting the given parameters into the formula, we have;
[tex]S = 0(7)+ \frac{1}{2}(2)(7^2)\\\\S = 0 + 49[/tex]
Distance, S = 49 meters.
Now, we can determine the net work done on the box:
[tex]Work\; done = Force[/tex] × [tex]distance[/tex]
[tex]Work\; done = 10[/tex] × [tex]49[/tex]
Work done = 490 Nm
Therefore, the net work done on the box is 490 Nm.
