lasie4. A 4 kg object is displaced to the right by a distance of 12 m underthe influence of the following forces: a 17 Nforce pu
lling to the right, a 36 N force that makesan angle of 30 deg with the horizontal-right.Calculate the net work done on the object. (1 point)A. O 462.747 JB. O1096.557 JC. O151.221 JD. O 675.066 J
1 answer:
The work done by a constant force in a rectilinear motion is given by:
where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:
Therefore, the total work done is 578.123 J and the answer is option E
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Answer:
The workdone is
Explanation:
From the question we are told that
The height of the cylinder is
The face Area is
The density of the cylinder is
Where is the density of freshwater which has a constant value
Now
Let the final height of the device under the water be
Let the initial volume underwater be
Let the initial height under water be
Let the final volume under water be
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as
So
=>
Now the work done is mathematically represented as
Substituting values
Answer:
The tension in the rope is 229.37 N.
Explanation:
Given:
Mass of the block is,
Coefficient of static friction is,
Angle of inclination is, °
Draw a free body diagram of the block.
From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.
Forces acting are and normal
. Now, there is no motion in the direction perpendicular to the inclined plane. So,
Consider the direction along the inclined plane.
The forces acting along the plane are and frictional force,
, down the plane and tension,
, up the plane.
Now, as the block is at rest, so net force along the plane is also zero.
Therefore, the tension in the rope is 229.37 N.
