Neddy Amendra
Forces in motion
2. Forces and equillibrium
1. Forces and acceleration
4. practise question
2. mass and weight
3. practise question
Forces and acceleration
force equals mass times acceleration.
A newton is the force required to accelerate a mass of 1kg by 1ms-2 in the direction of the force
Resultant force is the vector sum of all the forces
Mass weight and centre of mass
Mass of an object is the amount of matter in it . Weight is a force and is measured in Newtons.
A spacecraft is at rest in deep space and has a mass of 750 kg . The spacecraft fires its thrusters for 30 seconds producing a constant force of 1.5e3. Calculate the distance travelled in 30 seconds
f=mam=750kg t=30s u=0ms-1 a=f over m. 1.5e3 /750= 2ms-2 s=ut+1/2ate2 s=0(30)+(1/2 *2 *30e2) = 900.
next question
A smart phone of mass 100 grms slides down an inlcined table. The table is at an angle of 30 degrees from the horizontal . The phone experiences a constant friction of 0.3N. Calculate the magnitude of the net force on the phonein the plane parrallel to the tble. Give you answer to 2 sig figs
Fnetx = wx +friction(mg*siin(tita) +0.3 =(-0.1*9.81*sin(30)+0.3 =-0.1905 netx = 0.19N
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Neddy Amendra
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Transcript
Neddy Amendra
Forces in motion
2. Forces and equillibrium
1. Forces and acceleration
4. practise question
2. mass and weight
3. practise question
Forces and acceleration
force equals mass times acceleration.
A newton is the force required to accelerate a mass of 1kg by 1ms-2 in the direction of the force
Resultant force is the vector sum of all the forces
Mass weight and centre of mass
Mass of an object is the amount of matter in it . Weight is a force and is measured in Newtons.
A spacecraft is at rest in deep space and has a mass of 750 kg . The spacecraft fires its thrusters for 30 seconds producing a constant force of 1.5e3. Calculate the distance travelled in 30 seconds
f=mam=750kg t=30s u=0ms-1 a=f over m. 1.5e3 /750= 2ms-2 s=ut+1/2ate2 s=0(30)+(1/2 *2 *30e2) = 900.
next question
A smart phone of mass 100 grms slides down an inlcined table. The table is at an angle of 30 degrees from the horizontal . The phone experiences a constant friction of 0.3N. Calculate the magnitude of the net force on the phonein the plane parrallel to the tble. Give you answer to 2 sig figs
Fnetx = wx +friction(mg*siin(tita) +0.3 =(-0.1*9.81*sin(30)+0.3 =-0.1905 netx = 0.19N