When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
<h3>
Frictional force between the block and the horizontal surface</h3>
The frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Learn more about frictional force here: brainly.com/question/4618599
<span>B. velocity .................</span>
Answer:
25 N
Explanation:
Work is a product of force and perpendicular distance moved.
W=Fd where F is force exerted and d is perpendicular distance.
However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as 
Therefore, 
Making F the subject of the formula then
where 
is the angle of inclination. Substituting 190 J for W then 18 degrees for and 8 m for d then
Answer:
a) 4.485 kg b) 3.94 kg
Explanation:
since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .
44 = m * 9.81m/s^2
m = 44/9.81 = 4.485kg
b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)
44 = m (a +g)
44 = m (1.37 + 9.81)
44/11.18 = m
m = 3.94 kg
