A 1060 N crate is being pulled across a level floor by a force F of 300 N at an angle of 35° above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. Find the magnitude of the acceleration of the crate. answer in m/s2 if possible.
Thanks
see this example http://www.jiskha.com/display.cgi?id=1357972885
acceleration=net force/mass
To find the magnitude of the acceleration of the crate, we need to consider the forces acting on it. The forces at play are the applied force F, the force of gravity acting vertically downward, and the force of kinetic friction.
1. Resolve the applied force into horizontal and vertical components:
F_horizontal = F * cos(35°)
F_vertical = F * sin(35°)
2. Calculate the force of gravity:
F_gravity = m * g
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
3. Determine the force of kinetic friction:
F_friction = μ * F_normal
where μ is the coefficient of kinetic friction and F_normal is the normal force exerted by the floor on the crate. Since the crate is on a level floor, F_normal equals the force of gravity.
4. Substitute the given values into the equations:
F_horizontal = 300 N * cos(35°)
F_vertical = 300 N * sin(35°)
F_gravity = m * 9.8 m/s^2
F_friction = 0.25 * F_gravity
5. Rewrite Newton's second law equation:
F_net = m * a
where F_net is the net force acting on the crate and a is the acceleration we want to find.
6. Calculate the net force by summing the horizontal forces:
F_net = F_horizontal - F_friction
7. Substitute F_horizontal and F_friction into the equation:
F_net = (300 N * cos(35°)) - (0.25 * F_gravity)
8. Convert F_gravity into its equivalent mass:
F_gravity = m * 9.8 m/s^2
m = F_gravity / 9.8 m/s^2
9. Substitute m into the equation for F_net:
F_net = (300 N * cos(35°)) - (0.25 * (F_gravity / 9.8 m/s^2))
10. Finally, solve for the acceleration:
F_net = m * a
a = F_net / m
Now you can substitute the values you have and calculate the acceleration of the crate.