Question

Can someone please help me with these physics problems? I just don’t even know where to start.

Answer 1

#1

for the block of mass 5 kg normal force is given as

`F_n = mg`

`F_n = 5*9.8 = 49 N`

friction force is given as

`F_f = \mu F_n`

`F_f = 0.1*49 = 4.9 N`

Net force is given as

`F_(net) = ma`

`F_(net) = 5*2 = 10 N`

now we know that

`F_(net) = F_(app) - F_f`

`10 = F_(app) - 4.9`

`F_(app) = 14.9 N`

#2

Normal force is given as

`F_n = mg`

`F_n = 6*9.8`

`F_n = 58.8 N`

now we know that

`F_(net) = F_(app) - F_f`

`F_(net) = 0`

as object moves with constant velocity

`F_(app) = F_f = 15 N`

now for coefficient of friction we can use

`F_f = \mu F_n`

`15 = \mu * 58.8`

`\mu = 0.255`

#3

net force upwards is given as

`F = 1.2 * 10^(-4) N`

mass is given as

`m = 7 * 10^(-5) kg`

now as per newton's law we can say

`F = ma`

`1.2 * 10^(-4) = 7 * 10^(-5) * a`

`a = 1.71 m/s^2`

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

`F_(net) = F_(app) - F_f`

`ma = F_(app) - F_f`

`5*6 = 40 - F_(f)`

`F_f = 40 - 30 = 10 N`

part b)

for coefficient of friction we can use

`F_f = \mu F_n`

`10 = \mu * F_n`

here normal force is given as

`F_n = mg = 5*9.8 = 49 N`

now we have

`\mu = (10)/(49) = 0.204`

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

`d = v_i*t + (1)/(2)at^2`

`20 = 0 + (1)/(2)a(5^2)`

`a = 1.6 m/s^2`

now net force is given as

`F_(net) = ma`

`F_(net) = 10*1.6 = 16 N`

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

`a = (v_f - v_i)/(t)`

`a = (0 - 25)/(1.5) = -16.67 m/s^2`

now the force is gievn as

`F = ma`

`F = 5*16.67 = 83.3 N`