Posted by mary on Sunday, June 7, 2009 at 10:29pm.
F=22 N horizontal
d=0.73 m (displacement)
μ=0 (coefficient of friction)
The weight of the mass can be resolved into two components,
Normal to incline = mg cos(θ)
along incline (downwards) = mg sin(θ)
The horizontal force can be resolved into two components
Normal to incline = F sin(θ)
along incline (upwards) = F cos(θ)
Net force pushing mass up incline
= F cos(θ) - mg cos(θ)
= 22 cos(30) - 2.86*9.81 sin(30)
= 19.053 - 14.028
= 5.024 N up incline
Net acceleration up incline
= 5.024 / 2.86
= 1.757 m/s/s
initial velocity = 0
final velocity = v m/s
distance = d = 0.73 m
v2 - 02 = 2ad
physics - A horizontal force of magnitude 22 N is applied to a 3.0 kg book as ...
mary - The question is: a horizontal force of magnitude 22.0 N is applied to a 2...
Physics - a horizontal force of magnitude 18.0 N is applied to a 3.07 kg book as...
physices - A physics book slides off a horizontal table top with a speed of 1.30...
physics - a 1.50 kg book is sliding along a rough horizontal surface. At point A...
Physics - When a horizontal force of 350 is applied to a 86.0 box, the box ...
Physics - A student presses a book between his hands. The forces that he exerts ...
physics, - A student presses a book between his hands The forces that he exerts ...
physics - A student presses a book between his hands, as the drawing indicates. ...
Physics - A student presses a book between his hands, as the drawing indicates. ...
For Further Reading