Answer:
Explanation:
Given data:
Let the distance traveled by the object in the second case be 
In the given problem, work done by the forces are same in both the cases.
Thus,
Answer:
<h3>2,321.62Joules</h3>
Explanation:
The formula for calculating workdone is expressed as;
Workdone = Force * Distance
Get the force
F = nR
n is the coefficient of friction = 0.5
R is the reaction = mg
R = 46 ( 9.8)
R = 450.8N
F = 0.5 * 450.8
F = 225.4N
Distance = 10.3m
Get the workdone
Workdone = 225.4 * 10.3
Workdone = 2,321.62Joules
<em>Hence the amount of work done is 2,321.62Joules</em>
Answer:
25 m/s
Explanation:
Given that:
Initial speed, u = 4 m/s
Final velocity, V = 11 m/s
Time, t = 8 seconds
t2, = 16 seconds
Acceleration, a= (change in velocity) / time interval
a = (11 - 4) / 8
a = 7 / 8 = 0.875m/s²
Final velocity, v2 ;
Acceleration * t2
0.875 * 16 = 14
V2 = 14 m/s
Final speed : v + v2 = (11 + 14)m/s = 25m/s
The third hour was positive acceleration and the second hour was negative acceleration
