Fnety = (FT)(sin 32°) – Fg
Or the answer B, I checked it.
The work done on the barbell is -165.62 Nm.
Explanation:
Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.
In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.
Thus, 
Now, the force is -127.4 N and the displacement is 1.3 m.
So, 

So, the work done on the barbell is -165.62 Nm.
Answer:
Explanation:
The energy stored in the spring is used to throw the ball upwards . Let the height reached be h
stored energy of spring = 1/2 k y² , k is spring constant and y is compression created in the spring
stored energy of spring = potential energy of the ball
1/2 k y² = mgh , m is mass of the ball , h is height attained by ball
.5 k x .055² = .025 x 2.84
.0015125 k = .071
k = .071 / .0015125
= 46.9 N / m .
Answer:
18 times
Explanation:
According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.
In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.