Kinetic energy =1/2 mv^2

m=2ke/v^2 

m=2(34)/3.6^2 

m=5.24 

force normal = mg 
=5.24 x 9.8 
force normal = 51.4N

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5 0

3 years ago

A ball is dropped and falls with an acceleration of 9.8 m/s2 downward. It hits the ground with a velocity of 49 m/s downward. Ho

il63 [147K]

Hey!

NOTE-:

u= initial velocity
v= final velocity
g= acceleration due to gravity
t= time

u= 0
v= 49 m/s
t=?
g= 9.8 m/s^2

Using first equation of motion -

v-u=at
49-0= 9.8×t
49 = 9.8t
49/9.8= t
t= 5 second

Hope it helps...!!!

6 0

3 years ago

10. A 90 kg box is sliding across a surface at a constant velocity while experiencing a rightward applied

ANEK [815]

If the box is moving at constant velocity, net force must be zero, so:

F + fr = 0

fr = -F

fr = -40 N

4 0

3 years ago

Read 2 more answers

The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl

Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, $m_1=5\ kg$

Initial speed of block A, $u_1=5\ m/s$

Mass of block B, $m_2=8\ kg$

Initial speed of block B, $u_2=0$

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s$

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

8 0

3 years ago