Question

27. A block of mass 2 kg is resting on a horizontal
table. A force of 10 N is applied for 4 s on the
block in the horizontal direction. If g = 10 ms?,
and the cofficient of kinetic friction between the
block and the table -0.2. then the work done by
the net force is
(a) 144] (b)96) (c) 72J (d) 36J​

Answer 1

Answer:

72j

Explanation:

usework formula and force formula

Answer 2

Given Data :

time taken = 4 sec.

initial velocity = 0

mass of the block = 2 kg acceleration due to gravity = 10 m/s^2

here, Normal force 'N' = mg=2*10= 20 N

A/q

μ = friction ÷ Normal force

=> 0.2 = friction force ÷ N

=>0.2 * 20 N = friction force

Thus, we get force of friction = 4N

and, applied force = 10 N

then,

Net force = 10N-4N = 6N

Force = mass * acceleration

=> Force/mass = acceleration

=> 6N/2kg = acceleration

thus, we get accln. = 3 m/s^2

S = ut+1/2 at^2

= 0 + 1/2 *3 * 16

= 24 m

Now,

Work done = Fs

= 6N * 24 m

= 144 Joules