Explanation:
SI system is advantageous over mks system because SI system is intended to extend by its definition and has 7 base which is more than MKS system .
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Below is the solution:
9.43 m/s
<span>F = -kx </span>
<span>k = 400 N/m </span>
<span>PE= 0.5kx^2 = 0.5mv^2 </span>
<span>solve, v=9.43 m/s</span>
There are 4 forces acting on the block:
• its <u>w</u>eight (magnitude <em>w</em> = <em>m</em> <em>g</em> = 98 N, where <em>m</em> = 10 kg and <em>g</em> = 9.80 m/s²), pointed downward
• the <u>n</u>ormal force (mag. <em>n</em>), pointed upward
• <u>f</u>riction (mag. <em>f</em> = 15 N), pointing opposite the direction of motion
• the <u>p</u>ulling force (mag. <em>p</em> = 75 N), pointing at a 20° angle with the positive horizontal direction
Let the direction in which the block is being pulled be the positive horizontal direction, and upward the positive vertical direction. Split up each force into their horizontal and vertical components, then apply Newton's second law. We have a net horizontal force of
∑ <em>F</em> = <em>p</em> cos(20°) - <em>f</em> = <em>m</em> <em>a</em>
where <em>a</em> is the acceleration of the block. The net vertical force is 0, since the block gets dragged along the surface and doesn't move up or down (presumably).
Solve for <em>a</em> :
(75 N) cos(20°) - 15 N = (10 kg) <em>a</em>
<em>a</em> ≈ 5.4 m/s²
