Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.
Direction of the net force (β)
β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
Answer:
The system's potential energy is -147 J.
Explanation:
Given that,
Energy = 147 J
We know that,
System is isolated and it is free from external forces.
So, the work done by the external forces on the system should be equal to zero.
We need to calculate the system's potential energy
Using thermodynamics first equation
Put the value into the formula
Hence, The system's potential energy is -147 J.
Answer:
W = 18.88 J
Explanation:
Given that,
Constant force, F = 11.8 N (in +x direction)
Mass of an object, m = 4.7 kg
The object moves from the origin to the point (1.6i – 4.6j) m
We need to find the work is done by the given force during this displacement. The work done by an object is given by the formula as follows :
So, the work done by the given force is 18.88 J.
Answer:
9000 J
Explanation:
Convert minutes to seconds.
2 min = 120 s
Power = energy / time
75 W = E / 120 s
E = 9000 J
