(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
<h3>
Work done by the applied force</h3>
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
<h3>
Work done by the normal force</h3>
W = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
<h3>Work done force of gravity</h3>
The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
<h3> Work done by the net force on the block</h3>
∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
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<h3><u>Answer</u>;</h3>
= F0 L ( 1 - 1/e )
<h3><u>Explanation;</u></h3>
Work done is given as the product of force and distance.
In this case;
Work done = ∫︎ F(x) dx
= F0 ∫︎ e^(-x/L) dx
= F0 [ -L e^(-x/L) ] between 0 and L
= F0 L ( 1 - 1/e )
Answer:
D
Explanation:
6CO² + 6H²O > sunlight, chlorophyll, enzymes > C⁶H¹²O⁶ + 6O²
Answer:
Explanation:
Given
Two projectile is fired vertically upward
One has 4 times the mass of other
When Projectile is fired their trajectory is independent of mass of object. Also if they launched with same speed then both achieved same maximum height in same time and will hit the ground at the same moment.
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.
<span> U = (1/2)kx^2
</span><span> U = (1/2)(5.3)(3.62-2.60)^2
</span> U = <span>
<span>2.75706 </span></span>J
