Question

A 0.5 kg block is moving at constant speed on a horizontal surface. the friction force acting on the block is 20N, calculate uk

Answer 1

*uk is the coefficient of friction

Answer 2

would it just be .5 *20

Answer 3

any idea?

Answer 4

oh right

the opposing forces are equal as the speed is constant
friction al force = mN where N = normal reaction of ground on the block
= m* 0.5g = 20
m = 20 / (0.5*9.81)

Answer 5

hmmm - can't understand this - the uk must be in range 0 - 1.
maybe the values in question are wrong

Answer 6

so is your work right? whats its units?

Answer 7

i'm pretty sure the methods right - units???

Answer 8

cause the answer comes out to be 4.08? so 4.08 Neutons or what

Answer 9

not sure - ill check

Answer 10

ok?

Answer 11

no units as its a coefficient

Answer 12

so that answer is correct?

Answer 13

the answer is correct given the values - but in practice a force of 20 newtons is not possible for a 0.5 kg mass.

Answer 14

the question i typed is what it says on my paper...

Answer 15

frictional force = uk * normal reaction of ground on the block

normal reaction = mass * g = 0.5 * 9.81

Answer 16

now 2 N would be a possible value and uk would then be 0.408

Answer 17

can you explain why its .408

Answer 18

right
frictional force = 2 = 0.5*9.81 * uk
uk = 2 / (0.5*9.81) = 0.408

Answer 19

can you give me a medal too man?

Answer 20

of course if the mass is 5 kg then 20 N would be the force - that would give the same answer