A light spring having a force constant of 130N/m is used to pull a 7.50kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.50m/s^2?
Part A: By how much does the spring stretch if it pulls on the sled horizontally?
X=0.144m
Part B: By how much does the spring stretch if it pulls on the sled at 27.0∘ above the horizontal?
I figured out part a, but how do I find part b? Any advice?
Please and thanks!
http://answers.yahoo.com/question/index?qid=20110214141546AAp0rSP
To find the spring stretch when it pulls on the sled at an angle above the horizontal, you can use the concept of vector components. Here's how you can solve Part B:
1. Resolve the force applied by the spring into its horizontal and vertical components. Since the spring is pulling at an angle of 27.0° above the horizontal, the horizontal component is F_h = F * cos(27.0°) and the vertical component is F_v = F * sin(27.0°), where F is the force applied by the spring.
2. Calculate the net force acting horizontally. The net force acting on the sled horizontally is equal to the mass of the sled (m) multiplied by its acceleration (a). F_net_horizontal = m * a.
3. Equate the horizontal component of the force applied by the spring with the net force acting horizontally. F_h = F_net_horizontal.
4. Solve for the force applied by the spring, F. F = F_h / cos(27.0°).
5. Use Hooke's Law to find the spring stretch. The spring stretch (x) is related to the force applied by the spring by the equation F = k * x, where k is the force constant of the spring. Rearrange the equation to find x: x = F / k.
By substituting the value of F obtained in step 4 and the given value of k (force constant of the spring), you can calculate the spring stretch for Part B.
Note: Make sure to convert the angle from degrees to radians when using trigonometric functions (e.g., sin, cos) if your calculator uses radians.
Hope this helps!