Mechanics question help ?

A force of 6i+j N acts on a particle of mass 2 kg. The initial velocity of the particle is 2i−5j ms-1.

The velocity of the particle after 4 seconds is _i + ____j ms-1.

2 Answers
Mar 31, 2018

Answer:

See below.

Explanation:

#m(ddot x, ddot y) = (0,-mg)+(6,1)#

integrating once

#m(dot x,dot y)= m(v_x,v_y) = (0,-mg)t+(6,1)t+m(v_(0x),v_(0y))#

but #(v_(0x),v_(0y)) = (2,-5)# then

#(v_x, v_y) = (0,-g)t+1/m(6,1)t + (2,-5)# and after #4 # seconds

#(v_x, v_y) = 4 xx (0,-g)+4xx 1/m(6,1) + (2,-5)#

Apr 2, 2018

Answer:

#(14hati - 3hatj)\ "m/s"#

Explanation:

Force on particle is

#"F" = (6hati + hatj)\ "N"#

Acceleration of particle is

#"a" = "F"/"m" = ((6hati + hatj)\ "N")/"2 kg" = (3hati + 0.5hatj)\ "m/s"^2#

Velocity of particle after #"4 s"# is

#"v = u + at"#

#"v" = [(2hati - 5hatj)\ "m/s"] + [(3hati + 0.5hatj)\ "m/s"^cancel(2) × 4 cancel"s"]#

#"v" = (2hati - 5hatj)\ "m/s" + (12hati + 2hatj)\ "m/s"#

#"v" = (14hati - 3hatj)\ "m/s"#