Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
Answer:
24 N
Explanation:
= mass of the cube = 
Consider the three cubes together as one.
= mass of the three cubes together = 
= acceleration of the combination = 2 ms⁻²
= Force applied on the combination
Using Newton's second law

= Force by the left cube on the middle cube
Consider the forces acting on left cube, from the force diagram, we have

The correct answer is
<span>c. one person exerts more force than the other so that the forces are unbalanced.
In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:
</span>

<span>where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration.
In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.</span>
A. because I had this question yesterday.