Q.1 Find the force needed to accelerate a 2 kg block at 3m/s^2 up a rough plane (coefficient of friction 0.2) inclined at 25 degree to the horizontal if the force is

Question

a) Parallel to the slope
b) horizontal
c) at 45 degree to the upward verticle

Answer 1

First force is F which is pushing block upward
Second is the opposing forces, opposite to F
The block accelerates, so there is a net force (UNbalanced forces) which is net F =(F-opposing forces).
Since they are unbalanced and acceleration A results:

netF = m A
So
F-muR-mgsin25=ma
F-0.2mgcos25-mgsin25=ma
F-3.55-8.28=6
F=17.8N

b)
Fcos25-muR-19.6sin25=6.......eq.1
here R=Fsin25+mgcos25 because the force F is horizontal
Fcos25-muR-19.6sin25=6
substitute the value of R in eq.1
Fcos25-mu(Fsin25+mgcos25)-19.6sin25=6
0.906F-0.0845F-3.553-8.28=6
0.821F=17.833
F=17.833/8.21=21.7 N

about part 3 have no clue, i can not figure out the actual angle of force F

Answer 2

3. The angle which the force F makes with the direction of block motion (x-axis) is β=90 - 45 - 25 = 20 degr.
x: ma = F•cos β - m•g•sinα,
y: 0 = -m•g•cosα + N +Fsin β.

F = (m•a + m•g•sinα)/cos β = (2•3+2•9.8•sin25)/cos20 = 15.2 N

Answer 3

thanks a lot Elena, your help is much appreciated

Answer 4

No frictional force in no.3?