A block of mass 5kg is held at rest on a rough plane, inclined at 30degrees. The blocked is released.?

Question

The coefficient of friction is 0.2.

Show that the magnitude of the friction on the block is 8.5N?

Find the acceleration of the block?

Find the speed of the block when it has travelled 1.2 metres down the slope??

Answer 1

Resolving forces perpendicular to the plane:
N = mgcosx (where N = normal force)
N = 5(9.81)cos30
N = 42.5N

Frictional force = uN = (0.2)(42.5) = 8.5N (QED)

Resolving forces along the plane:
mgsinx - uN = ma
5(9.81)sin30 - 8.5 = 5a
a = 3.205m/s2

u=0m/s, a=3.205m/s2, s=1.2m, v=?
v^2 = u^2 + 2as
v^2 = 0 + 2(3.205)(1.2)
v = 2.77m/s

Answer 2

First of all outline your answers in sequential steps:
Draw a triangle without indicating the opposite side; this signifies the inclined plane.
Place the 5kg block on the hypoteneus, indicate the 30 degrees between the hypotenues and adjacent side.
Draw lines of force: one perpendicular to the top of the 5kg block with an arrow pointing upwards( put mgcos@ which is your N or reactional force.)
One perpendicular to the left side of the 5kg block as your P with an arrow pointing leftwards(put mgsin@ which is what's opposing friction).
One perpendicular to the right side of the 5kg block as your "f" which's your frictional force with the arrow parallel to the inclined plane and also close to it pointing rightwards to indicate that it's a frictional force acting.( put umgcos@, where "u" a.k.a "mu" is your coefficient of frictional force.
Lastly draw a normal line through the block and the inclined plane pointing downwards this indicates the weight.( put mg)
STEP2: show that the magnitude is 8.5N
f = umgcos@ where @ = 30 degrees, u = 0.2, m = 5kg and g = 9.8m/s^2 or 10m/s^2( acceleration due to gravity)
input: 0.2*5*9.8cos30
= 8.5N (proven)
STEP3: acceleration of the block
This tells you that initially the block was at rest; meaning that all the forces acting on the 5kg block where at equilibrum.
The part of the question that tells you the block was released means that there was now a resultant force acting on the 5 kg block that made it move from its static position. meaning that your equation will look like:
mgsin@ - umgcos@ = ma; where "ma" is that resultant force that moved it and the "a" is that acceleration being asked. So you make "a" the subject of the formula:
mg(sin@ - ucos@) = ma
("m" is common at both sides of the equality sign)
therefore;
g(sin@ - ucos@) = a
then you simply input:
9.8(sin30 - 0.2cos30)
9.8(0.5 - 0.2(0.866))
9.8(0.5 - 0.1732)
9.8(0.3268)
= 3.20264m/s^2 (

Answer 3

B. Block 2 at 60 ranges... because of the fact that the two blocks has the same mass and friction, you dont could understand the actuall hundreds, so "a" cancels out. the two blocks cant have the same speed because of the fact there are distinctive airplane inclines which cancels out "d". So the respond is block 2 because of the fact because of the fact the 60 degree airplane is tilted at an superior top than the 30 degree airplane, block 2 will slide down speedier via gravity and acceleration. purely attempt it with any merchandise and tilt one board bigger than the different, then you incredibly'll see...

Answer 4

simple , there is no self movement at 30 degrees as the lubricant between the two surfaces is non existant or, rather, not bought into the equation, neither is the providing force towards moving calculated, leaving only asumption of self movement of block as well as asumption of restraint force.'' HELD'' is not an accurate premise in this context, or the resistence factor of'' ROUGH''.therefore the question has not been established and cannot therefore have an answer

Answer 5

friction force= 0.2*5*9.8*cos30
=8.5N

use
mgsin30 - fmgcos30 = ma
get a from here as every term is known
now use

newton's third eqn of motion to get speed.

Answer 6

With your head under it.

Do your own Homework :)