The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
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Answer:
We know that potential energy of a body;
= mass(m)× gravitational acceleration(g) × height(h)
Lets find out the mass of the body
P.E. = mgh
=> 6500J = mass × 9.8m/s^2 × 12m
=>6500J = mass × ( 9.8 × 12 ) × ( m/s^2 × m)
=> 6500 Nm = m × 117.6 × m^2 / s^2
=> 6500/117.6 Ns^2/m = mass [°.° Ns^2/m = kg]
=> 55.272 Kg = mass
Therefore the mass of the body = 55.272 kg ~ <em>6</em><em>0</em><em> </em><em>k</em><em>g</em><em> </em>(Ans)
Hope it helps you
Out of the following choices given, hydraulic balances performs measurement in a closed compartment with no air currents to disturb measurement. The correct answer is C.
We can solve for the acceleration by using a kinematic equation. First we should identify what we know so we can choose the correct equation.
We are given an original velocity of 24 m/s, a final velocity of 0 m/s, and a time of 6 s. We and looking for acceleration (a) in m/s^2.
The following equation has everything we need:
So plug in the known values and solve for a:
0 = 24 + 6a
-24 = 6a
a = -4 m/s^2
Would it be an open cluster
