(a) The amount of work done by the horizontal force is 825 J.
(b) The amount of work done by gravitational force during the displacement is 514.1 J.
(c) The amount of work done by the normal force is 822.78 J
(d) The amount of work done by the frictional force is 90.5 J.
(e) The speed of the block after the displacement is 9.03 m/s.
The work done by the horizontal force is calculated as follows;
W = Fd
W = 150 x 5.5 = 825 J
W = mgh
W = mg(Lsinθ)
W = (18)(9.8)(5.5 x sin32)
W = 514.1 J
W = Fn x d
W = (mgcosθ) x d
W = (18)(9.8)(cos 32) x (5.5)
W = 822.78 J
W = Ff x d
W = μFn x d
W = μmgcosθ x d
W = (0.11)(18)(9.8)(cos32) x (5.5)
W = 90.5 J
∑F = ma
F - Ff = ma
F - μFn = ma
F - μmgcosθ = ma
150 - (0.11 x 18 x 9.8 x cos32) = 18a
150 - 16.456 = 18a
133.544 = 18a
a = 7.42 m/s²
v² = u² + 2as
v² = 0 + 2(7.42)(5.5)
v² = 81.62
v = √81.62
v = 9.03 m/s
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Answer:
A) 700 J
B) -514 J
C) 0 J
D) -139 J
E) 2.3 m/s
Explanation:
The amount of work done on an object is the product of applied force and displacement. The work done on an object will be translated to potential energy, kinetic energy, and heat (due to friction).
The horizontal force applied to the block is 150 N. The displacement of the block is (5.5 m)cos(32°) ≈ 4.664 m. The work done by the horizontal force is ...
Wh = F·d = (150 N)(4.664 m) ≈ 700. J
The force due to gravity is given by the equation ...
F = mg
Using g = -9.8 m/s² and the given mass, the gravitational force applied to the block is ...
F = (18 kg)(-9.8 m/s²) = -176.4 N
The upward displacement of the block is (5.5 m)sin(32°) ≈ 2.915 m. As above, the work is the product of this and the displacement. (Upward displacement is positive.)
Wg = F·d = (-176.4 N)(2.915 m) ≈ -514 J
The normal force on the block is perpendicular to the direction of motion. It will be the sum of the forces due to gravity and the horizontal force in the direction normal to the incline. In the direction into the incline, the force normal to the incline is ...
(150 n)sin(32°) +(176.4 N)cos(32°) ≈ 79.488 N +149.596 N ≈ 229.083 N
There is no displacement normal to the incline, so the work done by the normal force is ...
Wn = (229.1 N)(0 m) = 0 J
The friction force is the product of the coefficient of friction and the normal force:
F = µk·(229.083 N) = (0.11)(229.083 N) ≈ 25.20 N
This force operates in the direction opposite to the displacement. The work it does is ...
Wf = (-25.20 N)(5.5 m) ≈ -139 J
The total of all work done on the block will result in the block having a certain kinetic energy. That energy is ...
KE = (700 J) +(-514 J) +(0 J) +(-139 J) ≈ 46.92 J
This translates to the final velocity using the formula ...
v = √(2·KE/m) . . . . . m is mass
v = √(2·46.92 J/(18 kg)) ≈ 2.3 m/s
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Additional comments
• Per problem requirements, the results of calculations are shown to 3 (or 2) significant figures. Intermediate calculations are kept to a sufficient number of significant figures to ensure accuracy in all reported values.
• When displacement is in the direction opposite to the applied force, the work done by the force is negative.
• The attached diagram is intended to help visualization of the forces and their components.
