Senior 1 Physics An object with a mass of m 5kg moves horizontally to the left

1. When moving to the left, it is pulled to the right, and the object is decelerated in motion, and the direction of friction and tension is to the right

When the velocity decreases to 0:

VT2-VO2 2AS, F 30, A=10M S2100 20S

s=5m direction from the starting point to the left.

Elapsed time: vt=vo at

10＝10t

When t=1s is moving in the opposite direction, the frictional force is to the left: f is 30 , a=2m s2 when the velocity becomes 8m s:

vt=at8＝2t

t=4s displacement: vt2-vo2=2as

64＝4ss=16m

The starting point is at a velocity of 0.

Therefore, the whole time is 4 1 5s, and the displacement is: 16 5 11m

When the object moves to the left, the frictional force is to the right, that is, in the previous time, the force on the object is f+umg=50n, and the direction is to the right.

The velocity is 10m s, so, after t1 time, v=a*t1, t1=1s, the velocity of the object becomes 0, the direction of friction changes to the left, the force of the object becomes f-umg=10n, the direction is to the right, at this time a'=(f-umg) m=2m s2, the velocity should change to 8m s to the right, that is, v'=8m s, v'=a'*t2, therefore, t2=4s, the whole process time is t1+t2=5s

Total displacement s=s1-s2=1 2*(a*t1 squared-a.)'*t2 squared) = 1 2 * (10 * 1 - 2 * 16) = -11m, where the right direction is the positive direction.

Note that the displacement is a vector.

Are you done with this question...

(1) Force analysis of the object.

The support force of the ground towards the object n=mg-fsin

Friction f=un=u(mg-fsin).

The horizontal direction of the object is subjected to the combined external force fcos=fcos -u(mg-fsin)=maf(cos +sin)-umg=ma

a=f(cosθ+sinθ)/m-ug

*The rest is the math problem, find y=cos +sin max y 2 = 1 + sin2

Maximum when =45 degrees, y 2max=2

ymax=√2*//

Substitute the values and the maximum value of cos + sin 2 into a=f(cos + sin) m-ug

2) When the component of f in the vertical direction is balanced with the gravity of the object, the positive pressure between the object and the ground disappears, resulting in the disappearance of friction and the maximum acceleration of the object.

fsin53=mg

f=5mg/4=

a'=fcos53/m=3g/4=

The direction of friction is opposite to the direction of motion, to the right, and the force f is also to the right, then the resultant force ma= mg+f a=4m s 2 direction to the right.

The a here does not need to be added with a minus sign because it is said that the direction of acceleration is to the right, and the minus sign means that the direction of -a is to the right, so the a direction is to the left, which does not match the meaning of the title.

Friction = mg , opposite to the direction of motion.

Thrust to the right = 20n, opposite to the direction of motion.

So acceleration = (friction + thrust) m = g + 20n 10kg = 4m s and the direction is to the right.

First of all, the force analysis shows that the object receives a thrust f=20n to the right in the horizontal direction, and the friction force is opposite to the direction of motion, which is to the left, and the magnitude is umg=

f = f-f=, to the right, then the acceleration a=f = direction and the direction of the resultant force are the same, that is, to the right.

The acceleration of frictional force a=umg m=ug

Then the acceleration of the object a=f m-ug=20

In 0-2s, the acceleration of the object is (10-5) 2=, then the net force of the object is 5n, and the acceleration of the object in 2-6s is 10 4=, at this time there is no tension, there is only friction in the horizontal direction, so the friction force is 5n

So within 0-2s, call Hu f-f=ma, then f=10n handsome wolf hunting and troubleshooting Ying] team will answer for you in the bend.

The second half gives you the calculation of friction, and the reverse acceleration is, and the friction force is 5n.

In the first half of the blockage, the positive acceleration is the size of the object, and the resultant force is 5N, and the tensile force-friction force = the resultant force is 10

F=20n is decomposed into the horizontal direction of 16n to the left component and the vertical direction of 12n downward of the old component of the pants; Therefore, the total pressure on the ground is 22N, and the friction of sliding pure transport is f=11N; The frictional force and the 16n component to the left together have a total of 27n resultant force that hinders the movement of the luck and the acceleration of the deceleration is 27m s 2, and they stop and walk together within 1 s.

1. The work done by the pulling force is the same, because W=Fs, F is the same, S is the same 2, the work done by each force is different.

w total = wf + wf = the work done by friction plus the work done by the pulling force has no friction on a smooth horizontal plane, so the work done by friction is less, so the total work of the two is not the same.

Solution: w fs, both the pull force and the distance are the same.

w＝15n×

The object moves on the horizontal plane, the pulling force does positive work on the object, the frictional force does negative work on the object (or the object overcomes the frictional force to do the work), w total fs fs, the object moves on a smooth plane, the frictional force does the work is zero, then, the total work: w total fs

The object moves on a rough surface, frictional force: f mg

Total work: w total fs fs 15n

The work done by the pulling force is the same, the first one moves on the horizontal plane, and the work done by the pulling force is all converted into its kinetic energy.

The second type of motion on a rough surface, the work done by the pulling force is equal to the kinetic energy of the object plus the work done by the frictional force.

It is the same in (1) The work done by the tensile force is fx, f: the magnitude of the force, x: the distance the force moves (2) The position of the object changes is the same, so the total work of the object is also the same.

Sliding friction f= mg=

The tensile force, like the friction force of the chaotic shed, is to hinder the movement of the object to the right.