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A 925 N crate is being pushed across a level floor by a force of 325 N at an angle of 25 degrees above the horizontal. The coefficient of kinetic friction is 0.25. Find the magnitude of the acceleration of the crate.

Friday, July 23rd, 2010


First, let’s define what a Newton (N) is: “The Newton (N) is defined as the amount of force that, when acting on a 1 kg mass, produces an acceleration of 1 m/s/s (one meter per second per second). Therefore, 1 N = 1 kg =D7 1 m/s/s.”

The force of friction opposing the 325 N force is .25 * 925 N (which is equal to 231.25 N). The force pulling the crate is 325 N * cos(25), i.e. 294.55 N. (This is the component of force acting parallel to the floor.)

Take the difference between these two and you will have the net force pulling the crate (which is 63.3 N). Using Newton’s equation F = ma, you will find the acceleration by dividing the net force by the mass of the crate (the mass is 925 =/(one standard gravity [9.80665 m/s/s], or 94.32 kg). The acceleration is therefore 63.3 N / 94.32 kg, which equals 0.671 m/s/s.

This entry was posted on Friday, July 23rd, 2010 at 3:50 am and is filed under 9th Grade, Homework Answers, Physics Answers. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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