Question

A wooden block of mass = 500 is pressed against a spring of force constant = 1200 / and is held at rest at point , as shown in the figure. The compression in the spring is = 10 . When released, the block moves on a frictionless surface and climbs up a ramp of height ℎ = 80 . a) Find the speed of the block when it reaches to the top of the ramp at point . () When the block reaches the edge of the ramp, it leaves the ramp horizontally and undergoes a free-fall. b) Find the time of flight. c) Find the horizontal distance from the edge of the ramp where the block hits the ground. d) Find the speed of the block just before it hits the ground.

Answer 1

Answer:

The correct answer will be:

(a) 2.884 m/s

(b) 0.4 s

(c) 1.154 m

(d) 4.867 m/s

Explanation:

The given values are:

Mass, m = 500 g

on converting:

m = 0.5 kg

Force constant, k = 1200 N

Height, h = 0.8 m

x = 0.1 m

(a)...

As we know,

Kinetic Energy,

$KE=\frac{1}{2}kx^2 -mgh$

On putting the estimated values, we get

      $=\frac{1}{2}\times 1200\times (0.1)^2-0.5\times 9.8\times 0.8$

      $=6-3.92$

      $=2.08 \ J$

Now,

 $\frac{1}{2}mu^2=2.08$

 $\frac{1}{2}\times 0.5\times u^2=2.08$

 $u^2=8.32$

 $u=\sqrt{8.32}$

 $u=2.884 \ m/s$

(b)...

 $T=\sqrt{\frac{2h}{g} }$

On putting the values, we get

    $=\sqrt{\frac{2\times 0.8}{9.8}}$

    $=0.4 \ seconds$

(c)...

 $d=uT$

On putting the estimated values, we get

    $=2.884\times 0.4$

    $=1.154 \ m$

(d)...

 $V_{x}=u=2.884$

 $V_{y}=gT$

      $=9.8\times 0.4$

      $=3.92$

Now,

 $V=\sqrt{V_{x}^2+V_{y}^2}$

On putting the estimated values, we get

     $=\sqrt{(2.884)^2+(3.92)^2}$

     $=\sqrt{8.317456+15.3664}$

     $=\sqrt{23.684}$

     $=4.867 \ m/s$