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

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

Step-by-step 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)=(x)/(5),\\\frac{√(2)}{{2}}=(x)/(5),\\x=(5√(2))/(2)`

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

`(5√(2))/(2)=1.0\cdot a,\\a=(5√(2))/(2)\approx \boxed{3.5\:\mathrm{m/s^2}}`