Posted by **jane** on Thursday, June 18, 2009 at 7:43pm.

A 0.21 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by

x = (16 cm)cos[(16 rad/s)t + π/2 rad]

What force, applied to the block by the spring, results in the given oscillation?

why can you not just do F=ma?

## Answer This Question

## Related Questions

- physics please! - A 0.21 kg block oscillates back and forth along a straight ...
- physics - A 0.57 kg block oscillates back and forth along a straight line on a ...
- physics - The balance wheel of a watch oscillates with angular amplitude 1.7&#...
- physics - A block rests on a frictionless horizontal surface and is attached to ...
- CALCULUS PLEASE HELP!!! - SHOW WORK PLEASE!!! The displacement (in centimeters) ...
- physics ( please help me) - A 2.00kg block is pushed against a spring with force...
- physics - [20 pts] A 2.00 kg block is pushed against a spring with negligible ...
- physics - A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle ...
- trigonometry - calculate the linear velocity of an object rotating at an angular...
- Physics - An 8.00 kg block is placed on a rough horizontal surface and connected...

More Related Questions