# A box on a incline attached to a spring [closed]

I'm having a problem with a physics question from a exam I'm currently training for. I have no way of knowing if this is correct so I'm asking for help to review my answer!

This box is on a 20degree incline and the spring constant K is 2500 n/m. Before the box was attached to the spring the spring was 500mm and after 565 so a 65mm difference.

The question is what does the box weigh? DL= the difference of the spring before and after the box, 0.065 meters. K=2500n/ms

I thought abit about it and thought about hooks law which is F=DL K. Then I did this, MGsin20=DL K since the box is a 20 degree angle. then I just divided away g and sin 20 and came up with the box weighed 48.38kgs.

Please would appreciate some help if I'm in the right ballpark or not. Thanks for future answers!

## closed as off-topic by sammy gerbil, Jon Custer, Yashas, John Rennie, Kyle KanosMar 9 '17 at 11:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, Jon Custer, Yashas, John Rennie, Kyle Kanos
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Now all you have to do is equate that spring force by the force that is pulling the box down the slope - in the frictionless case that is just the plain old $\vec{F_g}$ times the sine of the slope angle.
• @Kennelmaster Yes, since $\vec{F_g} = m \vec{g}$ always pulls down, because that's where Earth is, the box also tries to move straight down. Since the slope is in the way, the box will move along the slope and that's where the sine reduces the force by an amount depending on the angle. – Wojciech Morawiec Mar 8 '17 at 7:54