Question

The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magnitude of the acceleration of the block?

Answer 1

Answer:

$3.5\:\mathrm{m/s^2}$

Explanation:

Newton's 2nd law is given as $\Sigma F = ma$.

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

$\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}$

Use this horizontal component of the force to solve for for the acceleration of the object:

$\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}$